LECTURES #5 & 6: HCP, POSITIONS,

DIRECTIONS, AND PLANES

1

ENGR 151: Materials of Engineering

2

Hexagonal Close-Packed Structure (HCP

– another view)

3

Hexagonal Close-Packed Structure (HCP

– top view)

• Coordination # = 12

4

HEXAGONAL CLOSE-PACKED STRUCTURE

(HCP)

• Coordination # = 12

• ABAB... Stacking Sequence

• APF = 0.74

• 3D Projection • 2D Projection

6 atoms/unit cell

ex: Cd, Mg, Ti, Zn

• c/a = 1.633

c

a

A sites

B sites

A sites Bottom layer

Middle layer

Top layer

ABAB... Stacking Sequence

THEORETICAL DENSITY, R

7

where n = number of atoms/unit cell

A = atomic weight

VC = Volume of unit cell = a3 for cubic

NA = Avogadro’s number

= 6.022 x 1023 atoms/mol

Density = =

VC NA

n A =

Cell Unit of Volume Total

Cell Unit in Atoms of Mass

THEORETICAL DENSITY, R

Ex: Cr (BCC)

A (atomic weight) = 52.00 g/mol

n = 2 atoms/unit cell

R = 0.125 nm

8

theoretical

a = 4R/ 3 = 0.2887 nm

actual

a R

= a 3

52.00 2

atoms

unit cell mol

g

unit cell

volume atoms

mol

6.022 x 1023

= 7.18 g/cm3

= 7.19 g/cm3

9

SUMMARY

• Atoms may assemble into crystalline or amorphous structures.

• We can predict the density of a material, provided we know the

atomic weight, atomic radius, and crystal geometry (e.g., FCC,

BCC, HCP).

• Common metallic crystal structures are FCC, BCC and HCP.

Coordination number and atomic packing factor are the same

for both FCC and HCP crystal structures.

• Interatomic bonding in ceramics is ionic and/or covalent.

SINGLE CRYSTALS

For a crystalline solid, when the periodic and

repeated arrangement of atoms extends

throughout without interruption, the result is

a single crystal.

The crystal lattice of the entire sample is

continuous and unbroken with no grain

boundaries.

For a variety of reasons, including the

distorting effects of impurities,

crystallographic defects and dislocations,

single crystals of meaningful size are

exceedingly rare in nature, and difficult to

produce in the laboratory under controlled

conditions. 10

Huge KDP (monopotassium phosphate)

crystal grown from a seed crystal in a

supersaturated aqueous solution at LLNL.

Below, silicon boule.

11

CRYSTALS AS BUILDING BLOCKS

• Some engineering applications require single crystals:

• Properties of crystalline materials

often related to crystal structure.

-- Ex: Quartz fractures more easily

along some crystal planes than

others.

-- diamond

Single crystals for abrasives

-- turbine blades

12

GLASS STRUCTURE

• Quartz is crystalline SiO2:

• Basic Unit: Glass is noncrystalline (amorphous)

• Fused silica is SiO2 - no impurities

have been added.

• Other common glasses contain

impurity ions such as Na+, Ca2+,

Al3+, and B3+

(soda glass)

Si 4+

Na +

O 2 -

Silicate glass tetrahedron SiO4

ANISOTROPY

The physical properties of single crystals of some substances depend on the crystallographic direction in which measurements are taken.

For example, the elastic modulus, electrical conductivity, and the index of refraction may have different values in the [100] and [111] directions.

The directionality of the properties is termed anisotropy and is associated with the atomic spacing.

13

ISOTROPIC

If measured properties are independent of the direction of measurement then they are isotropic.

For many polycrystalline materials, the crystallographic orientations of the individual grains are totally random.

So, though, a specific grain may be anisotropic, when the specimen is composed of many grains, the aggregate behavior may be isotropic.

14

15

POLYCRYSTALS

Most crystalline solids are composed of many small crystals

(also called grains).

Initially, small crystals (nuclei) form at various positions.

These have random orientations.

The small grains grow and begin to impinge on one another

forming grain boundaries.

16

Micrograph of a polycrystalline

stainless steel showing grains

and grain boundaries

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POLYCRYSTALS • Most engineering materials are polycrystals.

• Nb-Hf-W plate with an electron beam weld.

• Each "grain" is a single crystal.

• If grains are randomly oriented, overall component properties are not directional.

• Grain sizes typical range from 1 nm to 2 cm

(from a few to millions of atomic layers).

1 mm

Isotropic

Anisotropic

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POLYMORPHISM

Two or more distinct crystal structures for the same

material (allotropy/polymorphism)

titanium

, -Ti

carbon

diamond, graphite

BCC

FCC

BCC

1538ºC

1394ºC

912ºC

-Fe

-Fe

-Fe

liquid

iron system

19

SINGLE VS POLYCRYSTALS

• Single Crystals

-Properties vary with

direction: anisotropic.

-Example: the modulus

of elasticity (E) in BCC iron:

• Polycrystals

-Properties may/may not

vary with direction.

-If grains are randomly

oriented: isotropic.

(Epoly iron = 210 GPa)

200 mm

E (diagonal) = 273 GPa

E (edge) = 125 GPa

CRYSTAL SYSTEMS

20

7 crystal systems

14 crystal lattices

Unit cell: smallest repetitive volume that

contains the complete lattice pattern of a crystal.

a, b and c: lattice constants

The fourteen (14) types of Bravais lattices grouped in seven (7) systems.

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

Unit Cells Types

Primitive Face-Centered

Body-Centered End-Centered

A unit cell is the smallest component of the crystal that reproduces the whole

crystal when stacked together.

• Primitive (P) unit cells contain only a single lattice point.

• Internal (I) unit cell contains an atom in the body center.

• Face (F) unit cell contains atoms in the all faces of the planes composing the cell.

• Centered (C) unit cell contains atoms centered on the sides of the unit cell.

Combining 7 Crystal Classes (cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic,

trigonal) with 4 unit cell types (P, I, F, C) symmetry allows for only 14 types of 3-D lattice.

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

Lattice parameters in cubic, orthorhombic and hexagonal crystal systems.

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Cubic

Hexagonal

Tetragonal

Rhombohedral

Orthorhombic

Monoclinic

Triclinic

THE 7 TYPES OF CRYSTAL SYSTEMS

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Miller indices - A shorthand notation to describe certain crystallographic directions and planes in a material. Denoted by [ ], <>, ( ) brackets. A negative number is represented by a bar over the number.

Points, Directions and Planes in the Unit Cell

• Coordinates of selected points in the unit cell.

• The number refers to the distance from the origin in terms

of lattice parameters.

Point Coordinates

• Each unit cell is a reference or basis.

• The length of an edge is normalized as a unit of

measurement.

• E.g. if the length of the edge of the unit cell along the

X-axis is a, then ALL measurements in the X-direction

are referenced to a (e.g. a/2).

Point Coordinates – Contd.

Point coordinates for unit cell center are

a/2, b/2, c/2 ½ ½ ½

Point coordinates for unit cell corner are 111

Translation: integer multiple of lattice constants identical position in another unit cell

29

z

x

y a b

c

000

111

y

z

2c

b

b

Point Coordinates

EXAMPLE PROBLEMS

30

EXAMPLE PROBLEMS CONTD.

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EXAMPLE PROBLEMS CONTD.

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Determine the Miller indices of directions A, B, and C.

Miller Indices, Directions

(c) 2003 Brooks/Cole Publishing /

Thomson Learning™

GENERAL APPROACH FOR MILLER INDICES

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SOLUTION

Direction A

1. Two points are 1, 0, 0, and 0, 0, 0

2. 1, 0, 0, -0, 0, 0 = 1, 0, 0

3. No fractions to clear or integers to reduce

4. [100]

Direction B

1. Two points are 1, 1, 1 and 0, 0, 0

2. 1, 1, 1, -0, 0, 0 = 1, 1, 1

3. No fractions to clear or integers to reduce

4. [111]

Direction C

1. Two points are 0, 0, 1 and 1/2, 1, 0

2. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 1

3. 2(-1/2, -1, 1) = -1, -2, 2

2]21[ .4

(c) 2003 Brooks/Cole Publishing

/ Thomson Learning™

CRYSTALLOGRAPHIC DIRECTIONS

36

1. Vector repositioned (if necessary) to pass

through origin.

2. Read off projections in terms of

unit cell dimensions a, b, and c

3. Adjust to smallest integer values

4. Enclose in square brackets, no commas

[uvw]

ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]

-1, 1, 1

z

x

Algorithm

where overbar represents a

negative index

[ 111 ] =>

y

Problem 3.6, pg. 58

EXAMPLE PROBLEMS

37

EXAMPLE PROBLEMS, CONTD.

38

EXAMPLE PROBLEMS, CONTD.

39

Exam problems – make

sure you translate and

scale vector to fit WITHIN

unit cube if asked!

40

HCP CRYSTALLOGRAPHIC DIRECTIONS

Hexagonal Crystals

4 parameter Miller-Bravais lattice coordinates are

related to the direction indices (i.e., u'v'w') as follows.

=

=

=

' w w

t

v

u

) v u ( + -

) ' u ' v 2 ( 3

1 -

) ' v ' u 2 ( 3

1 - =

] uvtw [ ] ' w ' v ' u [

Fig. 3.8(a), Callister & Rethwisch 8e.

- a3

a1

a2

z

41

HCP CRYSTALLOGRAPHIC DIRECTIONS

Fig. 3.8(a), Callister & Rethwisch 8e.

- a3

a1

a2

z

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HCP CRYSTALLOGRAPHIC DIRECTIONS

1. Vector repositioned (if necessary) to pass

through origin.

2. Read off projections in terms of unit

cell dimensions a1, a2, a3, or c

3. Adjust to smallest integer values

4. Enclose in square brackets, no commas

[uvtw]

[ 1120 ] ex: ½, ½, -1, 0 =>

Adapted from Fig. 3.8(a),

Callister & Rethwisch 8e.

dashed red lines indicate

projections onto a1 and a2 axes a1

a2

a3

-a3

2

a 2

2

a 1

- a3

a1

a2

z

Algorithm

43

REDUCED-SCALE COORDINATE AXIS

44

PROBLEM 3.8

45

PROBLEM 3.8, CONTD.

46

PROBLEM 3.8, CONTD.

47

PROBLEM 3.8, CONTD.

48

PROBLEM 3.8, CONTD.

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PROBLEM 3.8, CONTD.

50

PROBLEM 3.9

51

PROBLEM 3.9, CONTD.

52

DRAWING HCP CRYSTALLOGRAPHIC

DIRECTIONS (I)

1. Remove brackets

2. Divide by largest integer so all values

are ≤ 1

3. Multiply terms by appropriate unit cell

dimension a (for a1, a2, and a3 axes)

or c (for z-axis) to produce

projections

4. Construct vector by placing tail at

origin and stepping off these

projections to locate the head

Algorithm (Miller-Bravais coordinates)

Adapted from Figure 3.10,

Callister & Rethwisch 9e.

53

DRAWING HCP CRYSTALLOGRAPHIC

DIRECTIONS (II)

Draw the direction in a hexagonal unit cell.

[1213]

4. Construct Vector

1. Remove brackets -1 -2 1 3

Algorithm a1 a2 a3 z

2. Divide by 3 1 3

1

3

2

3

1

3. Projections

proceed –a/3 units along a1 axis to point p

–2a/3 units parallel to a2 axis to point q

a/3 units parallel to a3 axis to point r

c units parallel to z axis to point s

 

[1 2 13]

p

q r

s

start at point o

Adapted from p. 72,

Callister &

Rethwisch 9e.

[1213] direction represented by vector from point o to point s

54

1. Determine coordinates of vector tail, pt. 1:

x1, y1, & z1; and vector head, pt. 2: x2, y2, & z2.

in terms of three axis (a1, a2, and z)

2. Tail point coordinates subtracted from head

point coordinates and normalized by unit cell

dimensions a and c

3. Adjust to smallest integer values

4. Enclose in square brackets, no commas,

for three-axis coordinates

5. Convert to four-axis Miller-Bravais lattice

coordinates using equations below:

6. Adjust to smallest integer values and

enclose in brackets [uvtw]

Adapted from p. 72, Callister &

Rethwisch 9e.

Algorithm

DETERMINATION OF HCP CRYSTALLOGRAPHIC

DIRECTIONS (II)

55

4. Brackets [110]

1. Tail location 0 0 0

Head location a a 0c

1 1 0 3. Reduction 1 1 0

Example a1 a2 z

1/3, 1/3, -2/3, 0 => 1, 1, -2, 0 => [ 1120 ]

6. Reduction & Brackets

Adapted

from p. 72,

Callister &

Rethwisch

9e.

DETERMINATION OF HCP CRYSTALLOGRAPHIC

DIRECTIONS (II)

Determine indices for green vector

2. Normalized

FAMILIES OF DIRECTIONS <UVW>

For some crystal structures, several

nonparallel directions with different

indices are crystallographically equivalent;

this means that atom spacing along each

direction is the same.

56

57

CRYSTALLOGRAPHIC PLANES

Adapted from Fig. 3.10,

Callister & Rethwisch 8e.

58

CRYSTALLOGRAPHIC PLANES

Miller Indices: Reciprocals of the (three) axial intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices.

Algorithm 1. Read off intercepts of plane with axes in terms of a, b, c 2. Take reciprocals of intercepts 3. Reduce to smallest integer values 4. Enclose in parentheses, no commas i.e., (hkl)

59

CRYSTALLOGRAPHIC PLANES z

x

y a b

c

4. Miller Indices (110)

example a b c z

x

y a b

c

4. Miller Indices (100)

1. Intercepts 1 1

2. Reciprocals 1/1 1/1 1/

1 1 0 3. Reduction 1 1 0

1. Intercepts 1/2

2. Reciprocals 1/½ 1/ 1/

2 0 0 3. Reduction 1 0 0

example a b c

60

CRYSTALLOGRAPHIC PLANES z

x

y a b

c

4. Miller Indices (634)

example 1. Intercepts 1/2 1 3/4

a b c

2. Reciprocals 1/½ 1/1 1/¾

2 1 4/3

3. Reduction 6 3 4

(001) (010),

Family of Planes {hkl}

(100), (010), (001), Ex: {100} = (100),

FAMILY OF PLANES

Planes that are crystallographically equivalent

have the same atomic packing.

Also, in cubic systems only, planes having the

same indices, regardless of order and sign,

are equivalent.

Ex: {111}

= (111), (111), (111), (111), (111), (111), (111), (111)

61

(001) (010), (100), (010), (001), Ex: {100} = (100),

_ _ _ _ _ _ _ _ _ _ _ _

FCC UNIT CELL WITH (110) PLANE

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BCC UNIT CELL WITH (110) PLANE

63

64

CRYSTALLOGRAPHIC PLANES (HCP)

In hexagonal unit cells the same idea is used

example a1 a2 a3 c

4. Miller-Bravais Indices (1011) (hkil)

1. Intercepts 1 -1 1 2. Reciprocals 1 1/

1 0

-1

-1

1

1

3. Reduction 1 0 -1 1

a2

a3

a1

z

Adapted from Fig. 3.8(b),

Callister & Rethwisch 8e.

65

DISTINCTION BETWEEN CRYSTALLOGRAPHIC

DIRECTIONS AND MILLER INDICES

Crystallographic Direction Crystallographic Planes

Associated with point

coordinates in 3D space

Associated with planar

structures in 3D space

Denoted by enclosing in

square brackets as [abc]

Denoted by enclosing in

regular parentheses (hkl)

EXAMPLE PROBLEMS

66

EXAMPLE PROBLEMS, CONTD.

67

Example of

family of

planes

EXAMPLE PROBLEMS, CONTD.

68

Don’t

forget to

slide plane

back within

unit cell!

EXAMPLE PROBLEMS, CONTD.

69

Strategy

Find answer in

3 coordinates

and then

convert to 4-

coordinate

representation

EXAMPLE PROBLEMS, CONTD.

70

71

ex: linear density of Al in [110]

direction

a = 0.405 nm

LINEAR DENSITY

Linear Density of Atoms LD =

a

[110]

Unit length of direction vector

Number of atoms

# atoms

length

3.5 atoms/nm a 2

2 LD = =

Adapted from Fig. 3.1(a),

Callister & Rethwisch 8e.

2 atoms per line, sharing computed

along vector length

72

PLANAR DENSITY OF (100) IRON Solution: At T < 912ºC iron has the BCC structure.

(100)

Radius of iron R = 0.1241 nm

R 3

3 4 a =

Adapted from Fig. 3.2(c), Callister & Rethwisch 8e.

2D repeat unit

= Planar Density = a 2

1

atoms

2D repeat unit

= nm2

atoms 12.1

m2

atoms = 1.2 x 1019

1

2

R 3

3 4 area

2D repeat unit

73

PLANAR DENSITY OF (111) IRON

Solution (cont): (111) plane 1 atom in plane/ unit surface cell

3 3 3

2

2

R 3

16 R

3

4

2 a 3 ah 2 area =

= = =

atoms in plane

atoms above plane

atoms below plane

a h 2

3 =

a 2

1

= = nm2

atoms 7.0

m2

atoms 0.70 x 1019

3 2 R 3

16 Planar Density =

atoms

2D repeat unit

area

2D repeat unit

74

X-RAY DIFFRACTION

Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation.

Can’t resolve spacings

Spacing is the distance between parallel planes of atoms.

X-RAY DIFFRACTION

75

Xray-Tube

Detector

Sample

Bragg´s Equation

d = λ/2 sinθ

d – distance between the same atomic planes λ – monochromatic wavelength θ – angle of diffracto- meter

Xray-Tube

Detector

Metal Target

(Cu or Co)

X-Ray Diffractometer

77

X-RAYS TO DETERMINE CRYSTAL STRUCTURE

X-ray intensity (from detector)

q

q c

d = n

2 sin q c

Measurement of

critical angle, qc,

allows computation of

planar spacing, d.

• Incoming X-rays diffract from crystal planes.

Adapted from Fig. 3.20,

Callister & Rethwisch 8e.

reflections must be in phase for a detectable signal

spacing between planes

d

q

q

extra distance travelled by wave “2”

CONSTRUCTIVE INTERFERENCE

Occurs at angle θ to the planes, if path length

difference is equal to a whole number, n, of

wavelengths (n equals order of reflection):

Bragg’s Law

BRAGG’S LAW

If Bragg’s Law is not satisfied then we get non-

constructive interference, which will yield a very

low-intensity diffracted beam.

For cubic unit cells:

BRAGG’S LAW

Specifies when diffraction will occur for unit

cells having atoms positioned only at the cell

corners

Atoms at different locations act as extra

scattering centers and results in the absence of

some diffracted beams.

81

X-RAY DIFFRACTION PATTERN

Adapted from Fig. 3.22, Callister 8e.

(110)

(200)

(211)

z

x

y a b

c

Diffraction angle 2q

Diffraction pattern for polycrystalline -iron (BCC)

Inte

nsity (

rela

tive

)

z

x

y a b

c

z

x

y a b

c

EXAMPLE PROBLEM

84

SUMMARY

• Atoms may assemble into crystalline or

amorphous structures.

• We can predict the density of a material, provided we

know the atomic weight, atomic radius, and crystal

geometry (e.g., FCC, BCC, HCP).

• Common metallic crystal structures are FCC, BCC, and

HCP. Coordination number and atomic packing factor

are the same for both FCC and HCP crystal structures.

• Crystallographic points, directions and planes are

specified in terms of indexing schemes.

Crystallographic directions and planes are related

to atomic linear densities and planar densities.

85

• Some materials can have more than one crystal

structure. This is referred to as polymorphism (or

allotropy).

SUMMARY

• Materials can be single crystals or polycrystalline.

Material properties generally vary with single crystal

orientation (i.e., they are anisotropic), but are generally

non-directional (i.e., they are isotropic) in polycrystals

with randomly oriented grains.

• X-ray diffraction is used for crystal structure and

interplanar spacing determinations.

ANNOUNCEMENTS

Homework 3 (Due Wed, 4/8/17)

3.20, 3.24, 3.29, 3.30, 3.32, 3.35

Quiz on Wednesday, March 1

Topic: Theoretical Density