The Science and Engineering of Materials, 4th edDonald R. Askeland – Pradeep P. Phulé

Chapter 4 – Imperfections in the Atomic and Ionic Arrangements

2

Objectives of Chapter 4

Introduce the three basic types of imperfections:

point defects, line defects (or dislocations), and

surface defects.

Explore the nature and effects of different types of

defects.

3

Chapter 4 Outline

4.1 Point Defects

4.2 Dislocations

4.3 Observing Dislocations

4.4 Schmid’s Law

4.5 Influence of Crystal Structure

4.6 Surface Defects

4.7 Importance of Defects

4

Point defects - Imperfections, such as vacancies, that are located typicallyat one (in some cases a few) sites in the crystal.

Extended defects - Defects that involve several atoms/ions and thus occurover a finite volume of the crystalline material (e.g., dislocations, grainboundaries, stacking faults, etc.).

Vacancy - An atom or an ion missing from its regular crystallographic site.

Section 4.1

Point Defects

5

Interstitial defect - A point defect produced when an atom is placed intothe crystal at a site that is normally not a lattice point.

Substitutional defect - A point defect produced when an atom is removedfrom a regular lattice point and replaced with a different atom, usually ofa different size.

6

Frenkel defect - A pair of point defects produced when an ion movesto create an interstitial site, leaving behind a vacancy.

Schottky defect - A point defect in ionically bonded materials. In orderto maintain a neutral charge, a stoichiometric number of cation andanion vacancies must form.

7

Determine the number of vacancies needed for a BCC iron crystal to have a

density of 7.87 g/cm3. The lattice parameter of the iron is 2.866 10-8 cm.

Example 4.1 SOLUTION

The expected theoretical density of iron can be calculated from the lattice

parameter and the atomic mass.

Example 4.1

Vacancy Concentrations in Iron

8

Example 4.1 SOLUTION (Continued)

Let’s calculate the number of iron atoms and vacancies that would be

present in each unit cell for the required density of 7.87 g/cm3:

Or, there should be 2.00 – 1.9971 = 0.0029 vacancies per unit cell. The

number of vacancies per cm3 is:

9

In FCC iron, carbon atoms are located at

octahedral sites at the center of each edge of

the unit cell (1/2, 0, 0) and at the center of

the unit cell (1/2, 1/2, 1/2).

In BCC iron, carbon atoms enter tetrahedral

sites, such as 1/4, 1/2, 0.

The lattice parameter is 0.3571 nm for FCC

iron and 0.2866 nm for BCC iron. Assume

that carbon atoms have a radius of 0.071 nm.

(1) Would we expect a greater distortion of

the crystal by an interstitial carbon atom

in FCC or BCC iron?

(2) What would be the atomic percentage of

carbon in each type of iron if all the

interstitial sites were filled?

Example 4.2

Sites for Carbon in Iron

10

Example 4.2 SOLUTION

1. We could calculate the size of the interstitial site

at the 1/4, 1/2, 0 location. The radius RBCC of the

iron atom is:

11

Example 4.2 SOLUTION (Continued)

The interstitial site in the BCC iron is smaller than the interstitial site in the

FCC iron. Although both are smaller than the carbon atom, carbon distorts

the BCC crystal structure more than the FCC crystal. As a result, fewer

carbon atoms are expected to enter interstitial positions in BCC iron than in

FCC iron.

For FCC iron, the interstitial site such as the 1/2, 0, 0 lies along

directions. Thus, the radius of the iron atom and the radius of the interstitial

site:

100

12

Example 4.2 SOLUTION (Continued)

2. In BCC iron: the number of tetrahedral interstitial sites is:

(24 sites)(1/2) = 12 interstitial sites per unit cell.

In FCC iron, the number of octahedral interstitial sites is:

(12 edges) (1/4) + 1 center = 4 interstitial sites per unit cell.

Because of the strain imposed on the iron crystal structure by the interstitial

atoms - particularly in BCC iron - the fraction of the interstitial sites that

can be occupied is quite small. The maximum atomic percentage of carbon

present in the two forms of iron under equilibrium conditions is

13

Section 4.2

Dislocations

Dislocation - A line imperfection in a crystalline material. Movement of

dislocations helps explain how metallic materials deform. Interference

with the movement of dislocations helps explain how metallic materials

are strengthened.

14

The perfect crystal (a) is cut and sheared one atom spacing, (b) and (c). The line along

which shearing occurs is a screw dislocation. A Burgers vector b is required to close a

loop of equal atom spacings around the screw dislocation.

Screw dislocation - A dislocation produced by skewing a crystal so that

one atomic plane produces a spiral ramp about the dislocation.

15

The perfect crystal in (a) is cut and an extra plane of atoms is inserted (b). The bottom

edge of the extra plane is an edge dislocation (c). A Burgers vector b is required to close a

loop of equal atom spacings around the edge dislocation.

Edge dislocation - A dislocation introduced into the crystal by adding an

‘‘extra half plane’’ of atoms.

16

A mixed dislocation. The screw dislocation at the front face of the crystal gradually

changes to an edge dislocation at the side of the crystal.

Mixed dislocation - A dislocation that contains partly edge components

and partly screw components.

17

When a shear stress is applied to the dislocation in (a), the atoms are displaced, causing the

dislocation to move one Burgers vector in the slip direction (b). Continued movement of the

dislocation eventually creates a step (c), and the crystal is deformed. (e) The motion of a

caterpillar is analogous to the motion of a dislocation.

18

Schematic of slip line, slip plane, and slip (Burgers) vector for (a) an edge dislocation and

for (b) a screw dislocation.

Slip - Deformation of a metallic material by the movement of dislocations

through the crystal.

19

Slip direction - The direction in the crystal in which the dislocation moves.

The slip direction is the same as the direction of the Burgers vector.

Slip plane - The plane swept out by the dislocation line during slip.

Normally, the slip plane is a close-packed plane, if one exists in the crystal

structure.

Slip system - The combination of the slip plane and the slip direction.

20

21

Schmid’s law - The relationship between shear stress, the applied stress, and

the orientation of the slip system - that is

(ø + λ) does not have to be 90°

Critical resolved shear stress - The shear stress required to cause a

dislocation to move and cause slip.

Section 4.4

Schmid’s Law

coscos

22

Section 4.5

Influence of Crystal Structure

Since different crystals or grains are oriented in different random directions, we

cannot apply Schmid’s law to predict the mechanical behavior of polycrystalline

materials.

Critical Resolved Shear Stress

In FCC metals, which have close-packed {111} planes, the critical resolved

shear stress is low.

BCC crystal structures contain no close-packed planes, and their critical

resolved shear stress is high.

In HCP metals such as zinc that have a c/a ratio greater than or equal to the

theoretical ratio of 1.633, the critical resolved shear stress is low, similar to

FCC metals.

In HCP titanium, the c/a ratio is less than 1.633; the close-packed planes are

spaced too closely together. Slip occurs on planes such as (1010), the “prism”

planes or faces of the hexagon, and the critical resolved shear stress is then as

great as or greater than in BCC metals.

23

Number of Slip Systems

HCP metals possess only one set of parallel close-packed planes, the

(0001) planes, and three close-packed directions, giving three slip

systems.

FCC metals contain four nonparallel close-packed planes of the form

{111} and three close-packed directions of the form 110 within each

plane, giving a total of twelve slip systems.

BCC metals have as many as 48 slip systems that are nearly close-packed.

Cross-slip - Consider a screw dislocation moving on one slip plane that

encounters an obstacle and is blocked from further movement. This

dislocation can shift to a second intersecting slip system, also properly

oriented, and continue to move.

In many HCP metals, no cross-slip can occur because the slip planes are

parallel (i.e., not intersecting).

Cross-slip is possible in both FCC and BCC metals because a number of

intersecting slip systems are present.

Cross-slip helps maintain ductility in FCC and BCC metals.

24

25

Section 4.6

Surface Defects Surface defects - Imperfections, such as grain boundaries, that form a two-

dimensional plane within the crystal.

Hall-Petch equation - The relationship between yield strength and grain size

in a metallic material—that is σy= σ0+Kd-1/2

σy: the yield strength (the level of stress necessary to cause a certain amount of

permanent deformation)

d: the average diameter of the grains

σ0 and K: constants for the metal.

26

The yield strength of mild steel with an average grain size of 0.05 mm is 20,000

psi. The yield stress of the same steel with a grain size of 0.007 mm is 40,000

psi. What will be the average grain size of the same steel with a yield stress of

30,000 psi? Assume the Hall-Petch equation is valid and that changes in the

observed yield stress are due to changes in dislocation density.

Example 4.3

Design of a Mild Steel

Example 4.3 SOLUTION

Thus, for a grain size of 0.05 mm the yield stress is

20 6.895 MPa = 137.9 MPa.

(Note:1,000 psi = 6.895 MPa). Using the Hall-Petch equation

27

Example 4.3 SOLUTION (Continued)

For the grain size of 0.007 mm, the yield stress is 40 6.895 MPa = 275.8 MPa.

Therefore, again using the Hall-Petch equation:

Solving these two equations K = 18.43 MPa-mm1/2, and σ0 = 55.5 MPa. Now we

have the Hall-Petch equation as

σy = 55.5 + 18.43 d-1/2

If we want a yield stress of 30,000 psi or 30 6.895 = 206.9 MPa, the grain size

will be 0.0148 mm.

28

ASTM grain size number (n) - A measure of the size of the grains in a

crystalline material obtained by counting the number of grains per

square inch a magnification of 100.

Microstructure of palladium (x100).

29

Suppose we count 16 grains per square inch in a photomicrograph taken at

magnification 250. What is the ASTM grain size number?

Example 4.4 SOLUTION

If we count 16 grains per square inch at magnification 250, then at

magnification 100 we must have:

N = (250/100)2 (16) = 100 grains/in.2 = 2n-1

Log 100 = (n – 1) log 2

2 = (n – 1)(0.301)

n = 7.64

Example 4.4

Calculation of ASTM Grain Size Number

30

Stacking Faults - A surface defect in metals caused by the improper

stacking sequence of closepacked planes.

Normally, a stacking sequence of ABC ABC ABC is produced in a perfect FCC

crystal. Suppose instead the following sequence is produced:

ABC ABAB CABC

In the portion of the sequence indicated, a type A plane replaces a type C plane.

This small region, which has the HCP stacking sequence instead of the FCC

stacking sequence, represents a stacking fault.

Stacking faults interfere with the slip process.

The ABCABCABC stacking

sequence of the FCC structure.

The ABABAB stacking

sequence of the HCP structure.

31

Twin Boundaries - A surface defect across which there is a mirror image

misorientation of the crystal structure. Twin boundaries can also move

and cause deformation of the material.

Twins can be produced when a shear force, acting along the twin boundary,

causes the atoms to shift out of position.

Twinning occurs during deformation or heat treatment of certain metals.

The twin boundaries interfere with the slip process and increase the strength

of the metal.

Application of a stress to the perfect crystal (a) may cause a displacement of the atoms, (b)

causing the formation of a twin. Note that the crystal has deformed as a result of twinning.

32

The high-energy grain boundaries are much more effective in blocking

dislocations than either stacking faults or twin boundaries.

33

Any imperfection in the crystal raises the internal energy at the location of the

imperfection because, near the imperfection, the atoms either are squeezed too closely

together (compression) or are forced too far apart (tension).

Effect on Mechanical Properties via Control of the Slip Process

Defects in materials, such as dislocations, point defects, and grain boundaries,

serve as “stop signs” for dislocations.

We can control the strength of a metallic material by controlling the number

and type of imperfections.

Section 4.7

Importance of Defects

34

Strain Hardening (work hardening) - Strengthening of a material by increasing

the number of dislocations by deformation, or cold working.

Solid-Solution Strengthening - Increasing the strength of a metallic material

via the formation of a solid solution.

A solid solution is formed when atoms or ions of a guest element or compound are

assimilated completely into the crystal structure of the host material.

Grain-Size Strengthening - Strengthening of a material by decreasing the grain

size. Grain boundaries resist dislocation motion, and thus, increasing the grain

boundary area leads to increased strength.

If the dislocation at point A moves to the

left, it is blocked by the point defect. If the

dislocation moves to the right, it interacts

with the disturbed lattice near the second

dislocation at point B. If the dislocation

moves farther to the right, it is blocked by

a grain boundary.

35

We would like to produce a bracket to hold ceramic bricks in place in a heat-

treating furnace. The bracket should be strong, should possess some ductility so

that it bends rather than fractures if overloaded, and should maintain most of its

strength up to 600oC. Design the material for this bracket, considering the

various crystal imperfections as the strengthening mechanism.

Example 4.5

Design/Materials Selection for a Stable Structure

36

Example 4.5 SOLUTION

In order to serve up to 600oC, the bracket should not be produced from

a polymer material. Instead, a metal or ceramic would be considered.

In order to have some ductility, dislocations must move and cause slip.

Because slip in ceramics is difficult, the bracket should be produced from a

metallic material.

We might add carbon to the iron as interstitial atoms or substitute

vanadium atoms for iron atoms at normal lattice points. These point defects

continue to interfere with dislocation movement and help to keep the strength

stable.

Of course, other design requirements may be important as well. For

example, the steel bracket may deteriorate by oxidation or may react with the

ceramic brick.