Chapter 3 -1 Theoretical Density, where n = number of atoms/unit cell A = atomic weight V C =...
-
Upload
virginia-flowers -
Category
Documents
-
view
221 -
download
1
Transcript of Chapter 3 -1 Theoretical Density, where n = number of atoms/unit cell A = atomic weight V C =...
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: