Material Engineering - Lecture - Deformation, Lecture notes of Material Engineering

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Deformation and

Strengthening

Mechanisms

Deformation and Strengthening Mechanisms

• (^) Plastic deformation caused by

dislocation movement.

• (^) Slip systems (slip plane, slip direction).
• (^) Resolved Shear Stress
• (^) Strengthening Mechanisms
• (^) Recovery, Recrystallization, Grain

Growth

Edge and Screw Dislocations

• (^) In an edge dislocation, localized lattice distortion exists along the end of an extra half-plane of atoms.
• (^) A screw dislocation results from shear distortion.
• (^) Many dislocations in crystalline materials have both edge and screws components; these are mixed dislocations.

• (^) Dislocation motion leads to plastic deformation.
• (^) An edge dislocation moves in response to a shear stress applied in a direction perpendicular to its line.
• (^) Extra half-plane at A is forced to the right; this pushes the top halves of planes B, C, D in the same direction.
• (^) By discrete steps, the extra 1/2-plane moves from L to R by successive breaking of bonds and shifting of upper 1/2-planes.
• (^) A step forms on the surface of the crystal as the extra 1/2-plane exits.

Dislocation Motion

• (^) The process by which plastic deformation is produced by dislocation motion is called slip (movement of dislocations).
• (^) The extra ½-plane moves along the slip plane.
• (^) Dislocation movement is similar to the way a caterpillar moves. The caterpillar hump is representative of the extra ½-plane of atoms.^7

Slip

When metals are plastically deformed, some fraction (roughly 5%) of energy is retained internally; the remainder is dissipated as heat. Mainly, this energy is stored as strain energy associated with dislocations. Lattice distortions exist around the dislocation line.

Slip Plane {111}: most dense atomic packing, Slip Direction ‹ 110 ›: highest linear density,

Slip System – FCC example

Stress and Dislocation Motion

• (^) Edge and screw dislocations move in response to shear stresses applied along a slip plane in a slip direction.
• (^) Even though an applied stress may be tensile, shear components exist at all but the parallel or perpendicular alignments to the stress direction.
• (^) These are resolved shear stresses (R).
• (^) Crystals slip due to resolved shear stress.

Condition for dislocation motion: crss R       cos cos  R Critical Resolved Shear Stress

• (^) In response to an applied tensile or compressive stress, slip (dislocation movement) in a single crystal begins when the resolved shear

stress reaches some critical value, crss.

• (^) It represents the minimum shear stress required to initiate slip and is a property of the material that determines when yielding occurs. (cos cos )max 13   crss y 

Deformation in a single crystal

• For a single crystal in tension, slip will occur along a number of equivalent and most favorably oriented planes and directions at various positions along the specimen.
• Each step results from the movement of a large number of dislocations along the same slip plane.

• (^) In addition to slip (dislocation movement), plastic deformation can occur by twinning.
• (^) A shear force can produce atomic displacements so that on one side of the plane (the twin boundary), atoms are located in mirror image positions to atoms on the other side.
• (^) Twinning may favorably reorient slip systems to promote dislocation movement. 16

Deformation by Twinning

Strengthening

• (^) The ability of a metal to deform plastically depends on the ability of dislocations to move.
• (^) Hardness and strength are related to how easily a metal plastically deforms, so, by reducing dislocation movement, the mechanical strength can be improved.
• (^) Greater mechanical forces will be required to initiate further plastic deformation.
• (^) To the contrary, if dislocation movement is easy (unhindered), the metal will be soft, easy to deform.

• (^) Grain boundaries are barriers to slip.
• (^) Barrier "strength“ increases with misorientation.
• (^) Smaller grain size: more barriers to slip.  yield   o  k y d  1 / 2

1. REDUCE GRAIN SIZE

Hall-Petch Equation: