Chapter 3 - 1

Theoretical Density,

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 VolumeTotal

Cell Unit in Atomsof Mass

Chapter 3: The Structure of Crystalline Solids

Chapter 3 - 2

• Ex: Cr (BCC)

A = 52.00 g/mol

R = 0.125 nm

n = 2 atoms/unit cell

theoretical

a = 4R/ 3 = 0.2887 nm

actual

aR

= a3

52.002

atoms

unit cellmol

g

unit cell

volume atoms

mol

6.022 x 1023

Theoretical Density,

= 7.18 g/cm3

= 7.19 g/cm3

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

Chapter 3 - 3

Densities of Material Classesmetals > ceramics > polymers

Why?

Data from Table B.1, Callister & Rethwisch, 8e.

(g

/cm

)3

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibers

Polymers

1

2

20

30Based 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

Graphite

Silicon

Glass -soda Concrete

Si nitride Diamond Al oxide

Zirconia

HDPE, PS PP, LDPE

PC

PTFE

PET PVC Silicone

Wood

AFRE*

CFRE*

GFRE*

Glass fibers

Carbon fibers

Aramid 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

Chapter 3 - 4

• 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. 8.33(c), Callister & Rethwisch 8e. (Fig. 8.33(c) courtesy of Pratt and Whitney).

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

Crystals as Building Blocks

Chapter 3 - 5

• 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).

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

1 mm

Polycrystals

Isotropic

Anisotropic

Chapter 3 - 6

• Single Crystals-Properties vary with direction: anisotropic.

-Example: the modulus of elasticity (E) in BCC iron:

Data from Table 3.3, Callister & Rethwisch 8e. (Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.)

• Polycrystals

-Properties may/may not vary with direction.-If grains are randomly oriented: isotropic. (Epoly iron = 210 GPa)-If grains are textured, anisotropic.

200 m Adapted from Fig. 4.14(b), Callister & Rethwisch 8e.(Fig. 4.14(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 vs PolycrystalsE (diagonal) = 273 GPa

E (edge) = 125 GPa

Chapter 3 - 7

Polymorphism

• Two or more distinct crystal structures for the same material.

• Allotropy-when found in elemental solids.   titanium

  , -Ti

carbon

diamond, graphite

BCC

FCC

BCC

1538ºC

1394ºC

912ºC

-Fe

-Fe

-Fe

liquid

iron system

Chapter 3 -

Tin (Its Allotropic Transformation)

8

White (or beta) tin, having a body-centered tetragonal crystal structure at room temperature, transforms, at 13.2°C (55.8°F), to gray (or alpha) tin, which has a crystal structure similar to diamond.

The rate at which this change takes place is extremely slow; however, the lower the temperature (below 13.2°C) the faster the rate. This transformation results in: Increase in volume (27 %), and, a decrease in density (from 7.30 g/cm3 to 5.77 g/cm3). Consequently, this volume expansion results in the disintegration of the white tin metal into a coarse powder of the gray allotrope. This produced some rather dramatic results in 1850 in Russia. The winter that year was particularly cold, and record low temperatures persisted for extended periods of time. The uniforms of some Russian soldiers had tin buttons, many of which crumbled because of these extreme cold conditions, as did also many of the tin church organ pipes. This problem came to be known as the “tin disease.”

Chapter 3 - 9

3.7   Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with the experimental value found inside the front cover.

3.8    Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4 g/cm3, and an atomic weight of 192.2 g/mol.

Chapter 3 -

Crystal Systems

10

Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal.

Fig. 3.4, Callister & Rethwisch 8e. a, b, and c are the lattice constant

Chapter 3 -

Crystal Structure and Crystal System

11

Q. What is the difference between crystal structure and crystal system?

A: A crystal structure is described by both the geometry of, and atomic arrangements within, the unit cell, whereas a crystal system is described only in terms of the unit cell geometry. For example, face-centered cubic and body-centered cubic are crystal structures that belong to the cubic crystal system.

Chapter 3 -

Point CoordinatesPoint coordinates for unit cell

center are

a/2, b/2, c/2 ½ ½ ½

Point coordinates for unit cell corner are 111

z

x

ya b

c

000

111

Chapter 3 -

Point coordinates for all atom positions for a BCC unit cell

13

Chapter 3 - 14

Crystallographic Directions

1. Vector repositioned (if necessary) to pass through origin.2. Read off projections in terms of unit cell dimensions a, b, and c3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvw]

ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]

-1, 1, 1

families of directions <uvw>

z

x

Algorithm

where overbar represents a negative index

[ 111 ]=>

y

Chapter 3 - 15

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 c3. Adjust to smallest integer values4. 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

a2

2

a1

-a3

a1

a2

z

Algorithm

Chapter 3 -

Problem 3.32

16

3.32 Determine the indices for the directions shown in the following cubic unit cell: