mse 3300-lecture note 05-chapter 03 the structure of crystalline solids 2

MSE 3300 / 5300 UTA Fall 2014 Lecture 5 -

Lecture 5. The Structure of Crystalline Solids (2)

Learning Objectives After this lecture, you should be able to do the following:

1. Given point coordinates within a unit cell, locate the point; given the location of a point, specify its point coordinates. 2. Given index integers for a direction within a unit cell, draw the direction; given a direction, specify its direction indices. 3. Given the indices for a plane within a unit cell, draw the plane; given a plane, specify its indices.

Reading • Chapter 3: The Structure of Crystalline Solids (3.8–3.17)

Multimedia • Virtual Materials Science & Engineering (VMSE):

http://www.wiley.com/college/callister/CL_EWSTU01031_S/vmse/

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Crystallographic Points, Directions, and Planes

• How to specify a particular point within a unit cell, a crystallographic

direction, or some crystallographic plane of atoms? • Use indices (e.g., three number) • Right-handed coordinate system

1. Points 2. Directions 3. Planes

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1. Points: Point Coordinates

• Specify a point (lattice position) within a unit cell

• Use three point coordinate indices: q, r, and s

• These indices are fractional multiples of a, b, and c unit cell edge lengths

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MSE 3300 / 5300 UTA Fall 2014 Lecture 5 - 4

Point Coordinates 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

z

x

y a b

c

000

111

y

z 2c

b

b


Crystallographic Directions

1. Determine coordinates of vector tail, pt. 1: x1, y1, & z1; and vector head, pt. 2: x2, y2, & z2. 2. Tail point coordinates subtracted from head point coordinates. 3. Normalize coordinate differences in terms of lattice parameters a, b, and c:

4. Adjust to smallest integer values 5. Enclose in square brackets, no commas

[uvw] ex: pt. 1 x1 = 0, y1 = 0, z1 = 0

=> 1, 0, 1/2

=> [ 201 ]

z

x

Algorithm

y

=> 2, 0, 1

pt. 2 head pt. 1:

tail

pt. 2 x2 = a, y2 = 0, z2 = c/2


Example Problem: Specification of Point Coordinate Indices

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Example Problem: Specification of Point Coordinate Indices

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2. Crystallographic Directions • Crystallographic direction is defined as a line directed between two points,

or a vector. 1. Construct a right-handed x-y-z coordinate system 2. Determine the coordinates of two points that lie on the direction vector (referenced to

the coordinate system)—Point 1 (vector tail): x1, y1, z1; Point 2 (vector head): x2, y2, z2 for the vector head

3. Subtract tail point coordinates from head point coordinates: x2-x1, y2-y1, and z2-z1. 4. Normalize (i.e., divide) the coordinate differences by using their respective lattice

parameters (a, b, and c): (x2-x1)/a, (y2-y1)/b, (z2-z1)/c. 5. If necessary, multiply or divide these three numbers by a common factor to reduce

them to the smallest integer values. 6. Enclose the three resulting indices in square brackets without comma: [uvw]. The u,

v, and w integers are the normalized coordinate differences referenced to the x, y, and z axes, respectively.

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Crystallographic Directions

-4, 1, 2

families of directions <uvw>

z

x

where the overbar represents a negative index

[ 412 ] =>

y

Example 2: pt. 1 x1 = a, y1 = b/2, z1 = 0 pt. 2 x2 = -a, y2 = b, z2 = c

=> -2, 1/2, 1

pt. 2 head

pt. 1: tail

Multiplying by 2 to eliminate the fraction


Crystallographic Directions: [100], [110], and [111] directions

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Determine the Indices for the Direction

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Determine the Indices for the Direction

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Construct a Specified Crystallographic Direction

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]011[Draw a direction with its tail located at the origin of the coordinate system.


Construct a Specified Crystallographic Direction

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]011[Draw a direction with its tail located at the origin of the coordinate system.


VMSE Screenshot – [101] Direction

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Directions in Hexagonal Crystals

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17

)2(31 VUu −=

)2(31 UVv −=

)( vut +−=

Ww =



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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.


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

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

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


Determine Directions in Hexagonal Crystals

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)2(31 VUu −=

)2(31 UVv −=

)( vut +−=

Ww =


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 [UVW] 5. Convert to four-axis Miller-Bravais lattice coordinates using equations below: 6. Adjust to smallest integer values and enclose in brackets [uvtw]


Algorithm

)2(31 VUu −= )2(

31 UVv −=

)( vut +−= Ww =

Determination of HCP Crystallographic Directions (ii)


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

5. Convert to 4-axis parameters

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

6. Reduction & Brackets


Determination of HCP Crystallographic Directions (ii)

Determine indices for green vector

2. Normalized


Determine the indices (four-index system) for the direction

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3. Crystallographic Planes

Adapted from Fig. 3.11, Callister & Rethwisch 9e.


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)


Crystallographic Planes z

x

y a b

c

4. Miller Indices (110)

example a b c z

x

y a b

c


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 2 0 0

example a b c


Crystallographic Planes z

x

y a b

c


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

“Family” of Planes: Crystallographically equivalent {hkl}

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


VMSE Screenshot – Crystallographic Planes

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Additional practice on indexing crystallographic planes


Crystallographic Planes (HCP) • In hexagonal unit cells the same idea is used


Crystallographic Planes (HCP) • In hexagonal unit cells the same idea is used

example a1 a2 a3 c

4. Miller-Bravais Indices (1011)

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



Determine the Miller-Bravais indices for the plane

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Summary of Equations

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Linear Density

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Linear Density (LD): Number of atoms per unit length whose centers lie on the direction vector for a specific crystallographic direction


ex: linear density of Al in [110] direction a = 0.405 nm

Linear Density

• Linear Density of Atoms ≡ LD =

a

[110]

Adapted from Fig. 3.1(a), Callister & Rethwisch 9e.

Unit length of direction vector Number of atoms

# atoms

length

1 3.5 nm a 2

2 LD − = =


Planar Density

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Planar Density (PD): Number of atoms per unit area that are centered on a particular crystallographic plane


Planar Density

37

Planar Density (PD): Number of atoms per unit area that are centered on a particular crystallographic plane


Crystallographic Planes

• We want to examine the atomic packing of crystallographic planes

• Iron foil can be used as a catalyst. The atomic packing of the exposed planes is important.

a) Draw (100) and (111) crystallographic planes for Fe. b) Calculate the planar density for each of these

planes.


Planar Density of (100) Iron Solution: At T < 912°C iron has the BCC structure.

(100)

Radius of iron R = 0.1241 nm

R 3

3 4 a =

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

Fig. 3.2(c), Callister & Rethwisch 9e [from W. G. Moffatt, G. W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 51. Copyright © 1964 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.]


Planar Density of (111) Iron Solution (cont): (111) plane 1 atom in plane/ unit surface cell

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

3 3 3 2

2

R 3

16 R 3

4 2 a 3 ah 2 area = = = =


VMSE Screenshot – Atomic Packing – (111) Plane for BCC

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• Single Crystals: The periodic arrangement of atoms extends throughout the entire specimen -Properties may vary with direction: anisotropic. -Example: the modulus of elasticity (E) in BCC iron:

Data from Table 3.4, Callister & Rethwisch 9e. (Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.)

• Polycrystals: Composed of a collection of many small crystals or grains; grain boundary -Properties may/may not vary with direction. -If grains are randomly oriented: isotropic. (Epoly iron = 210 GPa) -If grains are textured (have a preferential crystallographic orientation), anisotropic.

200 μm Adapted from Fig. 4.15(b), Callister & Rethwisch 9e. [Fig. 4.15(b) is courtesy of L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC (now the National Institute of Standards and Technology, Gaithersburg, MD).]

Single Crystals and Polycrystals E (diagonal) = 273 GPa

E (edge) = 125 GPa


Summary 1. Crystallographic points, directions, and planes 2. Point coordinates 3. Crystallographic directions 4. Crystallographic planes

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Homework 2 • 3.3, 3.8, 3.12, 3.19 • 3.26, 3.35, 3.41, 3.42, 3.48, 3.53, 3.56, 3.60 * Problems from Callister, 9th Edition

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Densities of Material Classes ρ metals > ρ ceramics > ρ polymers

Why?

Data from Table B.1, Callister & Rethwisch, 9e.

ρ (g

/cm

)

3

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibers Polymers

1

2

2 0

30 B ased on data in Table B1, Callister

*GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers

in an epoxy matrix). 10

3 4 5

0.3 0.4 0.5

Magnesium

Aluminum

Steels

Titanium

Cu,Ni Tin, Zinc

Silver, Mo

Tantalum Gold, W Platinum

G raphite Silicon Glass - soda Concrete

Si nitride Diamond Al oxide

Zirconia

H DPE, PS PP, LDPE PC

PTFE

PET PVC Silicone

Wood

AFRE * CFRE * GFRE* Glass fibers

Carbon fibers A ramid fibers

Metals have... • close-packing (metallic bonding) • often large atomic masses Ceramics have... • less dense packing • often lighter elements Polymers have... • low packing density (often amorphous) • lighter elements (C,H,O) Composites have... • intermediate values

In general


• Some engineering applications require single crystals:

• Properties of crystalline materials often related to crystal structure.

(Courtesy P.M. Anderson)

-- Ex: Quartz fractures more easily along some crystal planes than others.

-- diamond single crystals for abrasives

-- turbine blades Fig. 8.34(c), Callister & Rethwisch 9e. (courtesy of Pratt and Whitney)

(Courtesy Martin Deakins, GE Superabrasives, Worthington, OH. Used with permission.)

Crystals as Building Blocks


• 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 typically range from 1 nm to 2 cm (i.e., from a few to millions of atomic layers).

Fig. K, color inset pages of Callister 5e. (Courtesy of Paul E. Danielson, Teledyne Wah Chang Albany)

1 mm

Polycrystals

Isotropic

Anisotropic


X-Ray Diffraction

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

• Can’t resolve spacings < λ • Spacing is the distance between parallel planes of

atoms.


X-Rays to Determine Crystal Structure

X-ray intensity (from detector)

θ θ c

d = n λ 2 sin θ c

Measurement of critical angle, c, allows computation of planar spacing, d.

• Incoming X-rays diffract from crystal planes.


reflections must be in phase for a detectable signal

spacing between planes

d

θ λ

θ extra distance travelled by wave “2”


X-Ray Diffraction Pattern

Adapted from Fig. 3.22, Callister 8e.

(110)

(200)

(211)

z

x

y a b

c

Diffraction angle 2θ

Diffraction pattern for polycrystalline α-iron (BCC)

Inte

nsity

(rel

ativ

e)

z

x

y a b

c z

x

y a b

c

mse 3300-lecture note 05-chapter 03 the structure of crystalline solids 2

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