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The structure of Metals
1.
CHAPTER 3: The STRUCTURE ofMetals
ISSUES TO ADDRESS...
• How do atoms assemble into solid structures?
(for now, focus on metals)
• How does the density of a material depend on
its structure?
• When do material properties vary with the
sample (i.e., part) orientation?
Chapter 31
2.
Chapter 323.
Chapter 334.
structuressinglecrystal
polycrystal
noncrystal(amorphous)
• regularity longrange order
• hard sphere model ( Pauling’s model )
• three Dimensional (3D)
Chapter 34
5.
ENERGY AND PACKING• Non dense, random packing
Energy
typical neighbor
bond length
typical neighbor
bond energy
• Dense, regular packing
r
Energy
typical neighbor
bond length
r
typical neighbor
bond energy
Dense, regularpacked structures tend to have
lower energy.
Chapter 35
6.
MATERIALS AND PACKINGCrystalline materials...
• atoms pack in periodic, 3D arrays
• typical of: metals
many ceramics
some polymers
crystalline SiO2
Adapted from Fig. 3.18(a),
Callister 6e.
Noncrystalline materials...
• atoms have no periodic packing
• occurs for: complex structures
rapid cooling
"Amorphous" = Noncrystalline
Si
Oxygen
noncrystalline SiO2
Adapted from Fig. 3.18(b),
Callister 6e.
Chapter 36
7.
3.1 Introduction• Various types of atomic bonding
• Unit cell
• Three common crystal structures found in
metals
• Crystallographic points, directions, and
planes
Chapter 37
8.
Crystal Structures3.2 Fundamental Concepts
• A crystalline material is one in which the atoms
are situated in repeating or periodic array over
large atomic distances.
• Lattice means a threedimensional array of point
coinciding with atom positions.
Chapter 38
9.
3.3 Unit Cells• Crystal structure is often convenient to
subdivide the structure into a small repeat
entities called unit cells.
Chapter 39
10.
3.4 METALLIC CRYSTAL Structures• tend to be densely packed.
• have several reasons for dense packing:
Typically, only one element is present, so all atomic
radii are the same.
Metallic bonding is not directional.
Nearest neighbor distances tend to be small in
order to lower bond energy.
• have the simplest crystal structures.
We will look at three such structures...
Chapter 310
11.
SIMPLE CUBIC STRUCTURE (SC)• Rare due to poor packing (only Po has this structure)
• Closepacked directions are cube edges.
• Coordination # = 6
(# nearest neighbors)
(Courtesy P.M. Anderson)
Chapter 311
12.
ATOMIC PACKING FACTORAPF =
Volume of atoms in unit cell*
Volume of unit cell
*assume hard spheres
• APF for a simple cubic structure = 0.52
atoms
unit cell
a
R=0.5a
closepacked directions
contains 8 x 1/8 =
1 atom/unit cell
APF =
volume
atom
4
(0.5a)3
1
3
a3
volume
unit cell
Adapted from Fig. 3.19,
Callister 6e.
Chapter 312
13.
FACE CENTERED CUBICSTRUCTURE (FCC)
• Close packed directions are face diagonals.
Note: All atoms are identical; the facecentered atoms are shaded
differently only for ease of viewing.
• Coordination # = 12
Adapted from Fig. 3.1(a),
Callister 6e.
(Courtesy P.M. Anderson)
Chapter 313
14.
Chapter 31415.
ATOMIC PACKING FACTOR: FCC• APF for a FCC structure = 0.74
Closepacked directions:
length = 4R
= 2a
a
Adapted from
Fig. 3.1(a),
Callister 6e.
Unit cell contains:
6 x 1/2 + 8 x 1/8
= 4 atoms/unit cell
atoms
volume
4
3
( 2a/4)
4
unit cell
atom
3
APF =
volume
3
a
unit cell
Chapter 315
16.
FCC STACKING SEQUENCE• ABCABC... Stacking Sequence
• 2D Projection
A
B
B
C
A
B
B
B
A sites
C
C
B sites
B
B
C sites
• FCC Unit Cell
A
B
C
Chapter 316
17.
BODY CENTERED CUBICSTRUCTURE (BCC)
• Close packed directions are cube diagonals.
Note: All atoms are identical; the center atom is shaded
differently only for ease of viewing.
• Coordination # = 8
Adapted from Fig. 3.2,
Callister 6e.
(Courtesy P.M. Anderson)
Chapter 317
18.
4Ra
Chapter 318
19.
ATOMIC PACKING FACTOR: BCC• APF for a bodycentered cubic structure = 0.68
Closepacked directions:
length = 4R
= 3a
R
Adapted from
Fig. 3.2,
Callister 6e.
Unit cell contains:
1 + 8 x 1/8
= 2 atoms/unit cell
a
atoms
volume
4
3
( 3a/4)
2
unit cell
atom
3
APF =
volume
3
a
unit cell
Chapter 319
20.
HEXAGONAL CLOSEPACKEDSTRUCTURE (HCP)
• ABAB... Stacking Sequence
• 3D Projection
C
Adapted from Fig. 3.3,
• 2D Projection
A sites
Top layer
B sites
Middle layer
A sites
a
Bottom layer
Callister 6e.
• Coordination # = 12
• APF = 0.74
c/a=1.633
n=6=3+2×1/2+12×1/6
Chapter 320
1
21.
Chapter 32122.
Chapter 32223.
• Example Problem 3.2Show that the atomic packing factor for the FCC crystal structure is 0.74
Solution
The APF is defined as the fraction of solid sphere volume in a unit cell, or
APF
total sphere volume
V
S
total unit cell volume VC
Both the total sphere and unit cell volumes may be calculated in terms of
the atomic radius R. The volume for a sphere is 3 / 4 R 3 , and since there
are four atoms per FCC unit cell, the total FCC sphere volume is
16
4
VS 4 R 3 R 3
3
3
From Example Problem 3.1, the total unit cell volume is
VC 16 R 3 2
Therefore, the atomic packing factor is
VS 16 / 3 R 3
APF
0.74
3
VC
16 R 2
Chapter 323
24.
比較FCC
BCC
HCP
a ∞R
a=2√2R
a=4/√3R
a=2R
Atoms in unit
cell
Coordination
number
APF
4=8×1/8+6 2=1+8×1/8 6=3+2×1/2
×1/2
+12×1/6
12
8
12
0.74
0.68
0.74
Chapter 324
25.
3.5 DENSITY Computations,# atoms/unit cell
nA
VcNA
Volume/unit cell
(cm3/unit cell)
Atomic weight (g/mol)
Avogadro's number
(6.023 x 10 23 atoms/mol)
Example: Copper
Data from Table inside front cover of Callister (see next slide):
• 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 = 107cm)
Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 1023cm3
Result: theoretical Cu = 8.89 g/cm3
Compare to actual: Cu = 8.94 g/cm3
Chapter 325
26.
求 FCC 的 密Compare to actual: Cu = 8.94 g/cm3
Chapter 326
27.
Characteristics of Selected Elements at 20CAt. Weight
Symbol (amu)
Element
Aluminum Al
26.98
Ar
Argon
39.95
Ba
Barium
137.33
Be
Beryllium
9.012
B
Boron
10.81
Br
Bromine
79.90
Cd
Cadmium
112.41
Ca
Calcium
40.08
C
Carbon
12.011
Cs
Cesium
132.91
Cl
Chlorine
35.45
Chromium Cr
52.00
Co
Cobalt
58.93
Cu
Copper
63.55
F
Flourine
19.00
Ga
Gallium
69.72
Germanium Ge
72.59
Au
Gold
196.97
He
Helium
4.003
H
Hydrogen
1.008
Density
(g/cm3)
2.71
3.5
1.85
2.34
8.65
1.55
2.25
1.87
7.19
8.9
8.94
5.90
5.32
19.32

Crystal
Structure
FCC
BCC
HCP
Rhomb
HCP
FCC
Hex
BCC
BCC
HCP
FCC
Ortho.
Dia. cubic
FCC

Atomic radius
(nm)
0.143
0.217
0.114
Adapted from
Table, "Characof
0.149 teristics
Selected
0.197 Elements",
0.071 inside front
cover,
0.265 Callister 6e.
0.125
0.125
0.128
0.122
0.122
0.144
Chapter 327
28.
DENSITIES OF MATERIAL CLASSESmetals? ceramics? polymers
Why?
30
Ceramics have...
(g/cm3)
Metals have...
• closepacking
(metallic bonding)
• large atomic mass
• less dense packing
(covalent bonding)
• often lighter elements
Polymers have...
• poor packing
(often amorphous)
• lighter elements (C,H,O)
Composites have...
• intermediate values
Metals/
Alloys
20
Platinum
Gold, W
Tantalum
10
Silver, Mo
Cu,Ni
Steels
Tin, Zinc
5
4
3
2
1
0.5
0.4
0.3
Titanium
Aluminum
Magnesium
Graphite/
Ceramics/ Polymers
Semicond
Composites/
fibers
Based on data in Table B1, Callister
*GFRE, CFRE, & AFRE are Glass,
Carbon, & Aramid FiberReinforced
Epoxy composites (values based on
60% volume fraction of aligned fibers
in an epoxy matrix).
Zirconia
Al oxide
Diamond
Si nitride
Glasssoda
Concrete
Silicon
Graphite
Glass fibers
PTFE
Silicone
PVC
PET
PC
HDPE, PS
PP, LDPE
GFRE*
Carbon fibers
CFRE*
Aramid fibers
AFRE*
Wood
Data from Table B1, Callister 6e.
Chapter 328
29.
3.7 Crystal Systems• X,Y,Z: axes
• Lattice parameters: a, b, c:three edge lengths
α β γ : three interaxial angles
Chapter 329
30.
Chapter 33031.
6 lattice parameters方
簡單 方
體心 方
面心 方
方
正方
方
斜方晶系
單斜
三斜
Chapter 331
32.
3.8 Point CoordinatesChapter 332
33.
3.9 Crystallographic DirectionsA line between two points, or a vector
The steps are utilized in the determination of the three directional indices
1.A vector of convenient length is positional such that
it passes through the origin of the coordinate system.
Any vector may be translated throughout the crystal lattice
PROJECTION
without alteration, if parallelism is maintained.
2.The length of the vector projection on each of the three axes is determined:
these are measured in terms of the unit cell dimensions a,b,c.
3.These three numbers are multiplied or divided by a common factor to reduce
them to the smallest inter values.
4.The three indices, not separated by commas, are enclosed in square brackets,
thus: [uvw]. The u, v, and w integers correspond to the reduced projections along
the x, y, and z axes, respectively.
Chapter 333
34.
Miller indices[direction]
( plane )
Chapter 334
35.
Example Problem 3.6Determine the indices for the direction shown in the accompanying figure.
z
Projection on
x axis (a/2)
Projection on
y axis (b)
c
x
This procedure may be summarized as follows:
x
Projections
a/2
Projections (in terms of a, b, and c)
1/2
Reduction
1
Enclosure
y
y
b
1
2
[120]
z
0c
0
0
Chapter 335
36.
HEXAGONAL CRYSTALSA problem arises for crystals having hexagonal symmetry in that some crystallographic equivalent directions will not have the same set of indices. This is circumvented by utilizing a fouraxis, or MillerBravais, coordinate system as shown in
Figure 3.7. The three a1 , a2 , and a3 axes are all contained within a single plane
(called the basal plane), and at 120o angles to one another.
Chapter 336
37.
The z axis is perpendicular to this basal plane. Diretional indices, which are obtainedas described above, will be denoted by four indices,as [uvtw]; by convention, the
first three indices pertain to projections along the respective a1 , a2 , and a3 axes,z in
the basal plane.
Conversion from the threeindex system to the fourindex system,
[u’v’w’] [uvtw]
is accomplished by the following formulas:
n
(3.6a)
u ( 2u' v ' )
3
n
(3.6b)
v ( 2v ' u' )
3
(3.6c)
t ( u v )
(3.6d)
w nw'
Where primed indices are associated with the threeindex scheme and unprimed,
with the new MillerBravais fourindex system; n is a factor that may be required
to reduce u ,v , t , and w to the smallest integers. For example, using this conversion,
the [010] direction becomes [1210]. Several different directions are indicated in the
hexagonal unit cell (Figure 3.7a)
Chapter 337
38.
Example 3.8:• 試定出顯示於圖 (a)之 方單位晶胞的方向
指標。
決定方向向 在 a1、a2 和 z 軸的
投影 。其個別投影 分別為 a
a1 軸 a a2 軸 和 c z 軸
以單位晶胞 表示則變
成 1、1 和 1。因此
u’ = 1 v’ = 1 w’ = 1
Chapter 338
39.
將上面指標各乘以 3 得到u、v、t 和 w 分別為 1、1、2
和 3。因此 顯示於圖中之方
向為 [1123]
Chapter 339
40.
截距:1∞1倒 =101=(hkl)
i=(h+k)=(1+0)=1
(hkil)= 1011
Chapter 340
41.
3.10 Crystallographic PlanesThe procedure determine the h, k, l Miller index numbers
1.If the plane passes through the selected origin, either another parallel plane must be
constructed within the unit cell by an appropriate translation, or a new origin must be
established at the corner of another unit cell.
2.At this point the crystallographic plane either intersects or parallels each of the three
axes; the length of the planar intercept for each axis is determined in terms of the lattice
parameters a, b, and c.
3.The reciprocals of these numbers are taken. A plane that parallels an axis may be
considered to have an infinite intercept, and, therefore, a zero index.
4.If necessary, these three numbers are changed to the set of smallest integers by
multiplication or division by a common factor.
5.Finally, the integer indices, not separated by commas, are enclosed within parentheses,
thus: (hkl).
Chapter 341
42.
Chapter 34243.
Example Problem 3.9Determine the Miller indices for the plane shown in the accompanying
sketch (a)
x’ y’ z’
1 1/2
0 1 2
Chapter 343
44.
EXAM:3.10: 在一 方單位晶胞內畫出 011 的平面解
首先除去小括弧然後取倒 分別得到∞、 1 和 1
。此意味這特定平面平 於 x 軸 同時與 y 軸和 z
軸分別交截於 b 和 c 如附圖 a 所示。
011
Chapter 344
45.
Hexagonal crystals• Plane (hkil)
• i=  (h+k)
• a1 h a2 k a3 i
l z
截距:1∞1
倒 =101=(hkl)
i=(h+k)=(1+0)=1
(hkil)= 1011
Chapter 345
46.
Atomic ArrangementFCC
Chapter 346
47.
BCCChapter 347
48.
3.11 Linear and Planar Densities• LD:linear density
• LD=number of atom centered on direction
vector/length of direction vector
Chapter 348
49.
LD110 = 2 atoms/4R = 1/2RChapter 349
50.
PD = Planar density• PD = number of atoms centered on a plane/area
of plane
2
• PD110 = 2atoms/(4R) ×(2√2R)= 1/4R √2
(Fig3.10b)
Chapter 350
51.
3.12 Closedpacked crystal structuresChapter 351
52.
Chapter 35253.
Chapter 35354.
3.13 Single Crystalsf16_03_pg64 garnet 石榴石
Chapter 354
55.
SINGLE VS POLYCRYSTALS• Single Crystals
E (diagonal) = 273 GPa
Data from Table 3.3,
Callister 6e.
(Source of data is
R.W. Hertzberg,
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)
If grains are textured,
anisotropic.
Deformation and
Fracture Mechanics of
Engineering Materials,
3rd ed., John Wiley
and Sons, 1989.)
E (edge) = 125 GPa
200 m
Adapted from Fig.
4.12(b), Callister 6e.
(Fig. 4.12(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].)
Chapter 355
56.
3.14 Polycrystalline MaterialsChapter 356
57.
POLYCRYSTALS• Most engineering materials are polycrystals.
1 mm
Adapted from Fig. K,
color inset pages of
Callister 6e.
(Fig. K is courtesy of
Paul E. Danielson,
Teledyne Wah Chang
Albany)
• NbHfW plate with an electron beam weld.
• Each "grain" is a single crystal.
• If crystals are randomly oriented,
overall component properties are not directional.
• Crystal sizes typ. range from 1 nm to 2 cm
(i.e., from a few to millions of atomic layers).
Chapter 357
58.
DEMO: HEATING ANDCOOLING OF AN IRON WIRE
• Demonstrates "polymorphism"
Temperature, C
1536
The same atoms can
have more than one
crystal structure.
Liquid
BCC Stable
1391
longer
heat up
FCC Stable
914
Tc 768
BCC Stable
cool down
shorter!
longer!
magnet falls off
shorter
Chapter 358
59.
3.15 Anisotropyanisotropic
isotropic
Chapter 359
60.
3.16 Xray diffraction: Determination ofcrystal structures
Acrobat 文件
Chapter 360
61.
3.18constructive
destructive
Chapter 361
62.
3.19Bragg’s law
n = 2dhkl · sin
1.5402 Å
[Cu K ]
Chapter 362
63.
Bragg’s lawn = 2dhkl · sin
1.5402 Å
[Cu K ]
Chapter 363
64.
X光繞射和布 格定方體:
Chapter 364
65.
— diffraction techniques & apparatus‧diffractometer 繞射儀
power diffraction
diffraction angle ( 2θ)

xray , e beam , neutron beams used as a source
3.20
sample
xray generator
閘
Chapter 365
66.
determine : crystal structurecell dimension
(Laue method) crystal orientation (single crystal)
residual stress (compression strain)
crystal size estimation (Scherrer formula)
B (unit : rad )
2 B
0.9
補充 t B ×cos
B
BCC structure
FCC : all h , k , l odd or even ( 奇 or 偶 )
BCC : ( h + k +l ) must be even ( 偶 ) see Fig. 3.20
Chapter 366
67.
Figure3.21 Diffraction pattern for powdered lead.Chapter 367
68.
題 3.12 球平面間距 與繞射角對 BCC 鐵而言 計算(220)平面組之 (a)平面間距 與 (b)繞
射角。Fe 的晶格 為 0.2866nm。同時 假設所使用之單
光 射波長為 0.1790nm 且反射級 為 1。
解
(a) 平面間距dnkl之值可用 3.16 式決定 同時 a = 2.866nm
且h = 2 k = 2 及 l = 0 由於我們考慮的是 (220) 平面 因此
Chapter 368
P. 83
69.
(b) θ 值可以用 3.15 式 求得 由於是第一級反射 所以n = 1。
繞射角是 2θ 或
Chapter 369
P. 83
70.
example ()‧Fe(BCC) a=0.2866 , h,k,l = (2,1,1)
d hkl
a
2 1 1
2
2
2
0.117nm
λ=0.1542nm
‧Cu Kα1 xray
n‧λ = 2‧dhkl‧sinθhkl n=1
1 0.1542
sin 2 0.117 0.659
∴ 2 2 sin 1 82.44 o
Chapter 370
71.
3.17 Noncrystalline Solids3.22
Chapter 371
72.
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).
• Material properties generally vary with single
crystal orientation (i.e., they are anisotropic),
but properties are generally nondirectional
(i.e., they are isotropic) in polycrystals with
randomly oriented grains.
Chapter 372