Lecture 22: The mechanism of plastic deformation, part 2

1. Lecture 22: The mechanism of plastic deformation, part 2 PHYS 430/603 material Laszlo Takacs UMBC Department of Physics

2. Fracture • Brittle: There is very little plastic deformation preceding failure. In a uniform tensile sample it happens perpendicularly to the loading direction, induced by tension rather than shear. • Ductile: Substantial plastic deformation before failure. The direct cause is shear. Geometry depends on the material and conditions: • Slip at about 45° (or preferred slip direction) until failure (typical in single crystals.) • Necking to a point (very ductile polycrystals.) • “Cup and cone” in less ductile polycrystals. After some necking, failure starts inside and propagates at about 45° outward. Failure always happens much before the theoretical tensile strength is reached; it always begins at faults.

3. Technical materials always contain faults, e.g. microcracks. Why don’t they immediately result in failure? Create an elliptical microcrack in an infinite plate under uniform tension. • It releases energy due to decreased deformation: Uelastic = -c22/E • It requires surface energy Usurface = 4c Uelastic + Usurface has a maximum at ccrit = E/2 Microcracks smaller than this will heal rather than increase. Turned around, if a microcrack of length 2c exists inside the plate (or a notch of length c on the edge), crack propagation requires a minimum stress of crit = (E/c)1/2 2c

4. The primary mechanism of plastic deformation is slip due to dislocation motion. The required shear stress (Peierls stress): Which slip system is active depends on the crystal structure: Fcc: {1 1 1}<1 1 0>; often split into parallel partials. Hcp: {0 0 0 1}<1 1 -2 0> always available; several others, if c/a < 1.63. Bcc: {1 1 0}<1 1 1> is the best, but other slip planes with the same slip direction are close. More complex structures: Larger Burger’s vector makes slip difficult, material is usually brittle.

5. The interaction of dislocations • Dislocations interact via their elastic stress fields. Need to know: • Need to know the force acting on a dislocation due to a stress field • The stress field produced by a dislocation • Parallel dislocations repel, attract, shift each other • Dislocations on different slip planes must cut through each other

6. Work of external stress affecting the slip: W = ( l1 l3) bWork of force acting on the dislocation: W = (F l3) l1ComparedF =  b, where  is an external stress.In general geometryF = ( b) x s Peach-Koehler equation.(F is force per unit length.)

7. Except for a core about as wide as a single line of atoms, a dislocation can be represented with its elastic stress field. Edge dislocation: Strain is radial. Screw dislocation: Strain is parallel to the dislocation line. Strain goes to zero far from the dislocation line. With this conditions the stress field can be evaluated.

8. For example the stress field of an edge dislocation in the z direction is Here  is the asimuthal angle in cylindrical coordinates. Combining this with the P-K equation for parallel dislocations: The 45° lines are unstable, dislocation move away from there. The x component shows that dislocations in the same slip plane ( = 0°) repel each other, Fx = xyb >0. They form a train of dislocations. The y component aligns dislocations into small angle grain boundaries.

9. A general deformation requires that not all dislocations are parallel and they move across each other on different slip planes. This requires extra work; a dislocation always moves the most freely in a perfectly periodic lattice: • Crossing dislocations create jogs in the dislocation lines. (A jog is a step of the dislocation line out of the slip plane. Forming it requires energy.) • Some mobile dislocations contained in slip planes combine into a locked dislocation that is not mobile (Lomer lock).

10. The stages of strain hardening • Stage I: Dislocation density is low, dislocations move long distances along the primary slip plane without meeting an obstacle. • Stage II: Initially few dislocations exist in other slip systems, but they start to lead to cross-slips and locks, impeding dislocation motion. If dislocations are rendered immobile, new dislocations must form to continue the deformation. The dislocation density and the stress increase quickly. • Stage III: Cross slip of screw dislocations becomes important. It is a way to avoid obstacles and also results in the annihilation of some dislocations. The strain hardening rate gets smaller. The strain hardening rate can be characterized by  = d/d. The fastest strain hardening (in stage II) is about  = G/300 for most metals.

11. A dislocation can overcome an obstacle by increased shear stress alone, or thermal activation can help. Dislocation motion is easier at higher temperature, therefore the elastic limit is lower: • Forming metals is easier at high temperature. • Metals become weaker at high temperature At low temperature the elastic limit is high, a sample might break before plastic deformation begins, i.e. it becomes brittle.

12. The Frank - Reed source A single dislocation can provide a slip of b only. For macroscopic deformation many dislocations are needed, i.e. it is necessary to provide a mechanism for the generation of dislocations. Such a mechanism is the F-R source. Suppose a cross slip generates the dislocation segment BC. Without stress it is straight. Under stress it bows out to form an arc of radius R = Gb/2. As the stress increases, R decreases until 2R = BC = l is reached at 0 = Gb/l. At this point the arc becomes unstable, forms a closed loop and leaves the original line behind. This cycle can be repeated.

13. A Frank - Reed source in Si.Notice that the loops follow the structure of the lattice rather than being ideal circles.