chapter 3 crystalline structures part 1

7/30/2019 Chapter 3 Crystalline Structures Part 1

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Department of Mechanical Engineering-1-

Chapter 3

The Structure of Crystalline Solids

EGN 3365 Materials I
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Objectives of Chapter 3

Learn classification of materials based on atomic/ionic

arrangements

Describe the arrangements in crystalline solids based onlattice, basis, and crystal structure
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Issues to Address

How do atoms assemble into solid structures?

How do the structures of metals differ from those of

other materials?

How does the density of a material depend on its

structure?
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Types of Solids

Crystalline Material: Atoms self-organize in a periodic array

Single Crystal:

Atoms are in a repeating or periodic array over the entire

extent of the material Polycrystalline Material:

Comprised of many small crystals or grains

Amorphous Material:

Lacks atomic arrangement


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(c) 2003 Brooks/Cole Publishing / Thomson

Learning

Figure: Levels ofatomic arrangementsin materials: (a) NOORDER - Inertmonoatomic gaseshave no regularordering of atoms:(b,c) SHORT RANGEORDER (SRO) - Some

materials, includingwater vapor, nitrogengas, amorphous siliconand silicate glass haveshort-range order. (d)LONG-RANGE ORDER(LRO) - Metals, alloys,many ceramics andsome polymers have

regular ordering ofatoms/ions thatextends through thematerial.

Materials and Packing


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Single Crystals vs. Polycrystalline Materials


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Classification of Materials Based on Structure

Figure 3.4 (a) Photograph of a silicon single crystal.(b) Micrograph of a polycrystalline stainless steelshowing grains and grain boundaries (Co u r t e s y D r .M . H u a , D r . I . Ga r c i a , a n d D r . A . J. D e a r d o .)


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Figure: Liquid crystal display. Thesematerials are amorphous in one stateand undergo localized crystallization inresponse to an external electric fieldand are widely used in liquid crystaldisplays. (Co u r t e s y o f N i c k

K ou d i s / Ph o t o D is c / Ge t t y I m a g es .)

Classification of Materials Based on Structure


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Learning

Materials and Packing Summary


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Crystal Structure


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Definitions Lattice

Collection of points that divide space into smaller equally sizedsegments

Basis

Group of atoms associated with a lattice point

Unit cell

Subdivision of the lattice that still retains the overall characteristics ofthe entire lattice

Atomic radius

Apparent radius of an atom

Typically calculated from the dimensions of the unit cell, using close-packed directions (depends upon coordination number for metals,each atom has the same number of nearest-neighbor or touchingatoms)

Packing factor

The fraction of space in a unit cell occupied by atoms.


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Unit Cell


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-14- (c) 2003 Brooks/Cole Publishing / Thomson Learning

The fourteen

types of

lattices

grouped in

seven crystal

systems.


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Metallic Crystal Structures Metals are usually (poly)crystalline; although formation of

amorphous metals is possible by rapid cooling

As we learned in Chapter 2, the atomic bonding in metals isnon-directional no restriction on numbers or positions ofnearest-neighbor atoms large number of nearest neighbors

and dense atomic packing

Atomic (hard sphere) radius, R, defined by ion core radius -typically 0.1 - 0.2 nm

The most common types of unit cells are the faced-centeredcubic (FCC), the body-centered cubic (BCC) and the hexagonalclose-packed (HCP).


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Rare due to poor packing (only Po has this structure)

Close-packed directions are cube edges.

Coordination # = 6

(# nearest neighbors)

(Courtesy P.M. Anderson)

Simple Cubic Cell (SCC)


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APF for a simple cubic structure = 0.52

APF =a3

4

3 (0.5a)31

atoms

unit cellatom

volume

unit cell

volume

Atomic Packing Factor (APF)

APF =Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

Adapted from Fig. 3.23,

Callister 7e.

close-packed directions

a

R=0.5a

contains 8 x 1/8 =1 atom/unit cell


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Close packed directions are cube diagonals.--Note: All atoms are identical; the center atom is shaded

differently only for ease of viewing.

Body Centered Cubic Cell (BCC)

2 atoms/unit cell: 1 center + 8 corners x 1/8

Coordination # = 8

ex: Cr, W, Fe (), Tantalum, Molybdenum


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Atomic Packing Factor: BCC

a

APF =

4

3 ( 3a/4)32

atoms

unit cell atom

volume

a3 unit cell

volume

length = 4R =Close-packed directions:

3 a

APF for a body-centered cubic structure = 0.68

aR

Adapted fromFig. 3.2(a), Callister 7e.

a2

a3


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Close packed directions are face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded

differently only for ease of viewing.

Face Centered Cubic Cell (FCC)

4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8

Adapted from Fig. 3.1, Callister 7e.

ex: Al, Cu, Au, Pb, Ni, Pt, Ag

Coordination # = 12


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APF for a face-centered cubic structure = 0.74

Atomic Packing Factor: FCC

maximum achievable APF

APF =

4

3 ( 2a/4)34

atoms

unit cell atom

volume

a3

unit cell

volume

Close-packed directions:length = 4R = 2 a

Unit cell contains:6 x1/2 + 8 x1/8

=4 atoms/unit cella

2 a

Adapted from

Fig. 3.1(a),Callister 7e.


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Determine the relationship between the atomic radius and

the lattice parameter in SC, BCC, and FCC structures when

one atom is located at each lattice point.

Determining the Relationship between

Atomic Radius and Lattice Parameters


Learning


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We find that atoms touch along the edge of the cube in SC structures.

3

40

ra =

In FCC structures, atoms touch along the face diagonal of the cube.

There are four atomic radii along this lengthtwo radii from the face-

centered atom and one radius from each corner, so:

2

40

ra =

ra 20 =

In BCC structures, atoms touch along the body diagonal. There are two

atomic radii from the center atom and one atomic radius from each of

the corner atoms on the body diagonal, so


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A sites

B B

B

BB

B B

C sites

C C

CA

B

B sites

ABCABC... Stacking Sequence

2D Projection

FCC Unit Cell

FCC Stacking Sequence

B B

B

BB

B B

B sites

C C

CA

C C

CA

A

BC


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Example: Copper

crystal structure = FCC: 4 atoms/unit cell

atomic weight = 63.55 g/mol (1 amu = 1 g/mol)

atomic radius R = 0.128 nm (1 nm = 10 cm)-7

Compare to actual: Cu = 8.94 g/cm3

Result: theoretical Cu = 8.89 g/cm3

Theoretical Density


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Ex: Cr (BCC)A = 52.00 g/mol

R = 0.125 nm

n = 2

theoretical

a = 4R/ 3 = 0.2887 nm

actual

a

R

=a3

52.002

atoms

unit cell mol

g

unit cell

volume atoms

mol

6.023x1023

Theoretical Density,

= 7.18 g/cm3

= 7.19 g/cm3


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Determining the Density of BCC Iron

Determine the density of BCC iron, which has a lattice parameter of 0.2866

nm.

SOLUTION

Atoms/cell = 2, a0 = 0.2866 nm = 2.866 10-8 cmAtomic mass = 55.847 g/mol

Volume of unit cell = = (2.866 10-8 cm)3 = 23.54 10-24cm3/cell

Avogadros number NA

= 6.02 1023 atoms/mol

3

0a

3

2324/882.7

)1002.6)(1054.23(

)847.55)(2(

number)sadro'cell)(Avogunitof(volume

iron)ofmass)(atomicatoms/cellof(number

Density

cmg=

=

=


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metals ceramics polymers

Why?Metals have...

close-packing

(metallic bonding)

large atomic mass

Ceramics have... less dense packing

(covalent bonding)

often lighter elements

Polymers have... poor packing

(often amorphous)

lighter elements (C,H,O)

Composites have... intermediate values Data from Table B1, Callister 6e.

Densities of Material Classes
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chapter 3 crystalline structures part 1

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