Ch. 7 - Dislocations and Strengthening Mechanisms

Question 1

Edge dislocation

Answer 1

Localized lattice distortion exists among the end of an extra half-plane of atoms, which also defines the dislocation line.

Question 2

Screw dislocation

Answer 2

Results from a shear distortion; its dislocation line passes through the center of a spiral, atomic plane ramp.

Question 3

Slip

Answer 3

Process by which plastic deformation is produced by dislocation motion. Permanent plastic deformation results from the movement of dislocations in response to an applied shear stress.

Question 4

Slip plane

Answer 4

Crystallographic plane alone which the dislocation line traverses.

Question 5

Dislocation density

Answer 5

The total dislocation length per unit volume, of the number of dislocations that intersect a unit area of a random section.

10e3 for carefully molded crystals. 
10e9 for heavily deformed metals. 
10e5 for heat treated metals. 
10e2 for ceramic materials. 
0.1 for single silicon crystals used in integrated circuits.

Question 6

Lattice strain

Answer 6

Atomic lattice distortion exists around the dislocation line because of the extra half-plane of atoms. As a consequence, there are regions in which compressive, tensile, and shear lattice strains are imposed on the neighboring atoms.

Compressive strain above extra half-plane.
Tensile strain below extra half-plane.

Question 7

Slip plane

Answer 7

The preferred plane alone which slip occurs. This is the plane with the densest atomic packing.

Question 8

Slip direction

Answer 8

The direction of skip movement. This direction corresponds to the direction that is most closely packed with atoms in the most densely packed plane (ie: this direction has the highest linear density).

Question 9

Slip system

Answer 9

The combination of the slip plane and slip direction.

Question 10

Burger’s vector

Answer 10

This vector’s direction corresponds to a dislocation’s slip direction and its magnitude depends on the crystal structure.

Question 11

Resolved shear stresses

Answer 11

Edge, screw, and mixed dislocations move in response to shear stresses applied along the slip plane in the slip direction. Shear components exist at all alignments to the stress direction except parallel and perpendicular. These shear components are known as resolved shear stresses. Their magnitude depends on the applied stress and the orientation of both the slip plane and slip direction within the plane.

TR = (stress)cos(phi)cos(lambda)
Phi: angle between the normal vector of the slip plane and the direction of the force.
Lambda: angle between the slip and stress directions.

Question 12

Critical resolved shear stress

Answer 12

In response to some tensile or compressive stress, slip in a single crystal commenced on the most favorably oriented slip system when the resolved shear stress reaches some critical value. This value is the minimum shear stress required to initiate slip, and is a property of the material that determines when yielding occurs.

Yield stress = Critical resolved shear stress / (cos(phi)*cos(lambda))max

Question 13

Slip lines

Answer 13

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 length. Forms small steps on the surface of the single crystal that re parallel to one another and loop around the circumference of the specimen. Each step results from the movement of a large number of dislocations along the same slip plane.

Question 14

HCP, BCC, and FCC slip systems

Answer 14

HCP have few slip systems. This is why HCP metals are more brittle. No more than 6 slip systems.
FCC and BCC have a lot of slip systems. This is why FCC and BCC metals are so ductile. At least 12 slip systems.

Question 15

Plastic deformation of polycrystalline materials

Answer 15

Random crystallographic orientations of grains means that slip varies from one grain to the next. Dislocation motion occurs along slip system that has most favorable orientation.

During dislocation, grain boundaries do not usually come apart. Therefore, each individual grain is constrained in its distortion by neighboring grains. Grains usually go from equiaxial to stretched out.

Polycrystalline materials are stronger than single-crystal equivalents, so it takes more stress to initiate slip. Even though a single grain may be orientated favorably for slip, adjacent grains may not be.

Question 16

Twinning

Answer 16

A shear force can produce atomic displacements such that on one side of a plane (the twin boundary), atoms are located in mirror-image positions of the atoms on the other side.

Only occurs for a definite crystallographic plane in a specific direction that depends on crystal structure.

Regular slip deformation results in ledges / stair-steps. Twinning deformation results in a uniform wedge along the twin planes.

Occurs in BCC and HCP crystal structures at low temperatures and high rates of loading (shock loading) (ie: conditions under which slip is restricted)

Bulk plastic deformation is small in comparison to slip deformation. However, because it results in crystallographic reorientation, it may place new slip systems in positions that are favorable to the stress axis.

Question 17

Why do metals deform? How can we stop it?

Answer 17

The ability of a metal to plastically deform depends on the ability of the dislocations to move. All strengthening techniques for metals are based on restricting or hindering dislocation motion.

Question 18

How to reduce dislocation motion? (4)

Answer 18

Decrease grain size
Solid solution strengthening
Cold working / strain hardening
Precipitate hardening

Question 19

Strengthening by grain size reduction

Answer 19

Increases area of grain boundaries, which act as a barrier to dislocation motion because:

1. Grains are in different orientations. Dislocation passing from grain A to B will have to change its direction of motion.
2. Atomic disorder within grain boundary region will result in discontinuity of slip planes from grain A to B.

For high angle grain boundaries, dislocations tend to pile up, which introduces stress concentrations.

Grain size reduction also improves toughness as well as strength.

Question 20

Solid-solution strengthening

Answer 20

Metals can be strengthened and hardened by being alloyed with impurity atoms that go into either substitutional or interstitial solid solution.

Increasing concentration of impurity results in increase in tensile and yield strengths, but some ductility is lost.

Solid solution imposes lattice strains on surrounding host atoms. Solute atoms tend to diffuse and segregate around dislocations so as to reduce the overall strain energy. Cancel out some of the strain in the lattice surrounding the dislocation.

Question 21

Strain hardening / work hardening / cold working

Answer 21

Phenomenon whereby a ductile metal becomes harder and stronger as it is plastically deformed. Yield and tensile strength of material is increased, but some ductility is lost.

Dislocation density increases with cold work. Therefore, the average distance between dislocations decreases (ie: dislocations are closer together). Dislocation-dislocation strain interactions are repulsive. The net result is that the motion of a dislocation is hindered by the presence of other dislocations.

Question 22

Recovery

Answer 22

Some of the stores internal strain energy is relieved by virtue of dislocation motion as a result of enhanced atomic diffusion at the elevated temperature. This reduces the number of dislocations and dislocation configurations with low strain energies are produced. Properties previously altered are returned to normal, but grains are still in a relatively high energy state.

Question 23

Recrystallization

Answer 23

The formation of a new set of strain-free and equiaxed grains that have low dislocation densities and are characteristic of the precold-worked condition. Metal becomes softer, weaker, yet more ductile.

New grains form as very small nuclei and grow until they completely consume the parent material. Involves short-range diffusion.

The degree or fraction of recrystallization increases with time.

Question 24

Grain growth

Answer 24

After recrystallization, the strain-free grains continue to grow if the metal specimen is left at the elevated temperature. Occurs by the migration of grain boundaries. Not all grains can enlarge, but large ones grow at the expense of small ones that shrink, so the average grain size increases with time. Boundary motion is just the short range diffusion of atoms from one side of the boundary to the other.