Metals- Dislocation Sources, Partial Dislocations, Crystal Deformation Flashcards
How are dislocation generated in a perfect crystal?
Homogeneous nucleation of dislocations from a perfect crystal. This occurs at very very high stresses when sufficient energy is present to cause shear of the perfect crystal (most difficult mechanism)
Other ways of generating dislocations
From interfaces that essentially act as regions of stress concentration which induces dislocation generation. These could be: Interfaces with other phases Grain boundaries Migrating boundaries The free surface of a crystal
How do Frank-Read sources work?
Start with a dislocation pinned at either side which experiences a force τb when stress is applied to the material. Because it’s pinned it will bow out to balance the line tension 1/2 Gb^2 with the force acting on the dislocation. When the two segments at the bottom of the loop touch each other they annihilate as their line vectors are in opposite directions (so opposite b). The dislocation loop formed may continue to propagate. Dislocation segment between the two pinning joints may go on to produce further dislocations
What does a perfect dislocation do?
Moves a plane of atoms one full position
Why do dislocations often dissociate into two dislocations?
It is often energetically favourable for them to dissociate into two dislocations with smaller Burgers vectors. This requires the passage of two smaller dislocations that are not lattice vectors, and have Burgers vectors b2 and b3, known as partial dislocations
Burgers vectors of partial dislocations in fcc metal
b=a/6 <11-2>
What is the stacking fault?
A change in the stacking sequence. Can be due to action of the first partial dislocation
Proving that formation of partial dislocations is favourable in fcc metals
Need to show b1^2 greater than b2^2+b3^2.
Perfect dislocation has b1=a/2 [-101]
Partial dislocations have b2=a/6 [-211] and b3=a/6 [-1-12]
Find square the a terms and times by sum of squares in brackets.
Find b1^2=a^2/2
b2^2+b3^2=a^2/3 so the reaction is favourable
Stacking faults energy
The energy penalty associated with the creation of the stacking fault
Répulsion of partial dislocations
When a perfect dislocation splits into two partial dislocations the elastic strain fields from each of them will repel each other (same sign on same plane)
Partial dislocation repulsion balanced with stacking fault energy
For high stacking fault energies the dislocations are closer together and vice versa
Can partial dislocations cross slip?
No. They must recombine into the perfect dislocation before they can cross slip
How does stacking fault energy affect strengthening?
Cross slip of partials hardest in materials with low stacking fault energy because the partials are most widely separated. For materials with low stacking fault energy, work-hardening is observed to occur rapidly. Presence of stacking faults themselves also increases the strength of the material
Why do stacking faults increase the strength of the material?
Due to the change in the stacking of the atomic planes, strain fields develop around the section of the crystals with the “missing” plane. This train field further inhibits the strength of the material
Shear stress vs shear strain graph for fcc single crystal deformation
Initial elastic region. Then plastic deformation. Stage I is horizontal. Stage II curves up to roughly diagonal. Stage III curve decreases in gradient until horizontal
Stage I of plastic deformation
The horizontal section. Known as easy glide. One operative slip system. As deformation advances tensile axis rotated towards slip direction. Schmid factor constantly changing on every possible slip system.
Stage II of plastic deformation
The curve up to about diagonal section. Tensile axis has rotated into a position in which two slip systems share the largest Schmid factor. So two operative slip systems known as duplex slip. Dislocation interactions that produce jogs, locks and pile ups. Leads to an increase in critical resolved shear stress (strength of material) required to move further dislocations on these slip systems. Process of increasing the strength is known as work hardening or strain hardening
Stage III of plastic deformation
Resolved shear stress becomes sufficiently high to activate more slip systems and allow dislocations to bypass obstacles. Transition to Stage III correlates with the staking fault energy in fcc metals.
How does transition to Stage III depend on stacking fault energy for fcc metals?
Cross slip occurs more readily in metals with higher stacking fault energies. Means transition to Stage III in high SFE energy materials occurs at lower applied stresses. Opposite true for low SFE materials (delayed transition and more work hardening)
Stress vs strain graph for polycrystalline metals
Normal shape with straight elastic region then dome
Why is there no Stage I deformation in polycrystalline metals?
Due to each grain experiencing duplex slip and work hardening at different points during deformation. Each grain has a different orientation in comparison to the tensile axis so exhibit different Schmid factors
Schmid factor of randomly oriented polycrystalline samples
About 1/3
Taylor factor
τcrss roughly σY x 1/3
So σY roughly 3τcrss
This is for polycrystalline samples
Why is strength of polycrystalline sample much higher than what Taylor factor predicts?
Due to presence of grain boundaries and the requirements for grains to be compatible with each other, that can constrain deformation in some respects
