CHAPTER 3: The STRUCTURE of Metals 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 3-1

Chapter 3-2

Chapter 3-3

structures

single-crystal poly-crystal non-crystal(amorphous)

• regularity  long-range order • hard sphere model ( Pauling’s model ) • three- Dimensional (3-D) Chapter 3-4

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, regular-packed structures tend to have lower energy. Chapter 3-5

MATERIALS AND PACKING Crystalline 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 3-6

3.1 Introduction • Various types of atomic bonding • Unit cell • Three common crystal structures found in metals • Crystallographic points, directions, and planes

Chapter 3-7

Crystal Structures 3.2 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 three-dimensional array of point coinciding with atom positions.

Chapter 3-8

3.3 Unit Cells

• Crystal structure is often convenient to subdivide the structure into a small repeat entities called unit cells.

Chapter 3-9

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 3-10

SIMPLE CUBIC STRUCTURE (SC) • Rare due to poor packing (only Po has this structure) • Close-packed directions are cube edges. • Coordination # = 6 (# nearest neighbors)

(Courtesy P.M. Anderson)

Chapter 3-11

ATOMIC PACKING FACTOR APF =

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 close-packed 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 3-12

FACE CENTERED CUBIC STRUCTURE (FCC) • Close packed directions are face diagonals. --Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.

• Coordination # = 12

Adapted from Fig. 3.1(a),

Callister 6e.

(Courtesy P.M. Anderson)

Chapter 3-13

Chapter 3-14

ATOMIC PACKING FACTOR: FCC • APF for a FCC structure = 0.74 Close-packed 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 3-15

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 3-16

BODY CENTERED CUBIC STRUCTURE (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 3-17

4R

a

Chapter 3-18

ATOMIC PACKING FACTOR: BCC • APF for a body-centered cubic structure = 0.68 Close-packed 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 3-19

HEXAGONAL CLOSE-PACKED STRUCTURE (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 3-20 1

Chapter 3-21

Chapter 3-22

• Example Problem 3.2 Show 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 3-23

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 3-24

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 = 10-7cm) Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10-23cm3

Result: theoretical Cu = 8.89 g/cm3 Compare to actual: Cu = 8.94 g/cm3 Chapter 3-25

求 FCC 的理論密度

Compare to actual: Cu = 8.94 g/cm3

Chapter 3-26

Characteristics of Selected Elements at 20C At. 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, "Charac-----of 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 3-27

DENSITIES OF MATERIAL CLASSES metals? ceramics? polymers Why?

30

Ceramics have...

 (g/cm3)

Metals have... • close-packing (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 Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers in an epoxy matrix). Zirconia Al oxide Diamond Si nitride Glass-soda 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 3-28

3.7 Crystal Systems

• X,Y,Z: axes • Lattice parameters: a, b, c:three edge lengths α，β，γ : three interaxial angles

Chapter 3-29

Chapter 3-30

6 lattice parameters 立方

簡單立方 體心立力方 面心立方

六方

正方 菱方

斜方晶系

單斜

三斜 Chapter 3-31

3.8 Point Coordinates

Chapter 3-32

3.9 Crystallographic Directions A 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 3-33

Miller indices [direction] ( plane )

Chapter 3-34

Example Problem 3.6 Determine 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 3-35

HEXAGONAL CRYSTALS A 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 four-axis, or Miller-Bravais, 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 3-36

The z axis is perpendicular to this basal plane. Diretional indices, which are obtained as 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 three-index system to the four-index 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 three-index scheme and unprimed, with the new Miller-Bravais four-index 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 3-37

Example 3.8:

• 試定出顯示於圖 (a)之六方單位晶胞的方向 指標。

決定方向向量在 a1、a2 和 z 軸的 投影量。其個別投影量分別為 a （a1 軸），a（a2 軸）和 c（z 軸 ），以單位晶胞參數來表示則變 成 1、1 和 1。因此： u’ = 1 v’ = 1 w’ = 1

Chapter 3-38

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 3-39

截距:1∞1 倒數=101=(hkl) i=-(h+k)=-(1+0)=-1 (hkil)= 1011 Chapter 3-40

3.10 Crystallographic Planes The 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 3-41

Chapter 3-42

Example Problem 3.9 Determine the Miller indices for the plane shown in the accompanying sketch (a)

x’ y’ z’  -1 1/2 0 1 2 Chapter 3-43

EXAM:3.10: 在一立方單位晶胞內畫出 011的平面 解 首先除去小括弧然後取倒數，分別得到∞、－1 和 1 。此意味這特定平面平行於 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 3-44

Hexagonal crystals • Plane (hkil) • i= - (h+k) • a1－h，a2－k，a3－i ，l－z

截距:1∞1 倒數=101=(hkl) i=-(h+k)=-(1+0)=-1 (hkil)= 1011

Chapter 3-45

Atomic Arrangement

FCC

Chapter 3-46

BCC

Chapter 3-47

3.11 Linear and Planar Densities

• LD:linear density • LD=number of atom centered on direction vector/length of direction vector

Chapter 3-48

LD110 = 2 atoms/4R = 1/2R

Chapter 3-49

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 3-50

3.12 Closed-packed crystal structures

Chapter 3-51

Chapter 3-52

Chapter 3-53

3.13 Single Crystals

f16_03_pg64 garnet 石榴石

Chapter 3-54

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 3-55

3.14 Polycrystalline Materials

Chapter 3-56

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)

• Nb-Hf-W 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 3-57

DEMO: HEATING AND COOLING 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 3-58

3.15 Anisotropy

anisotropic

isotropic

Chapter 3-59

3.16 X-ray diffraction: Determination of crystal structures

Acrobat 文件

Chapter 3-60

3.18

constructive

destructive

Chapter 3-61

3.19

Bragg’s law n = 2dhkl · sin   1.5402 Å [Cu K]

Chapter 3-62

Bragg’s law n = 2dhkl · sin   1.5402 Å [Cu K]

Chapter 3-63

X-光繞射和布拉格定律 立方體:

Chapter 3-64

— diffraction techniques & apparatus ‧diffractometer 繞射儀 power diffraction diffraction angle ( 2θ) -

x-ray , e beam , neutron beams used as a source 3.20

sample x-ray generator 閘

Chapter 3-65

determine : crystal structure cell 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 偶數 ) BCC : ( h + k +l ) must be even ( 偶數 ) see Fig. 3.20

Chapter 3-66

Figure3.21 Diffraction pattern for powdered lead.

Chapter 3-67

例題 3.12 球平面間距離與繞射角 對 BCC 鐵而言，計算(220)平面組之 (a)平面間距離與 (b)繞 射角。Fe 的晶格參數為 0.2866nm。同時，假設所使用之單 光輻射波長為 0.1790nm，且反射級數為 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 3-68

P. 83

(b) θ 值可以用 3.15 式來求得，由於是第一級反射，所以 n = 1。

繞射角是 2θ 或

Chapter 3-69

P. 83

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 x-ray  n‧λ = 2‧dhkl‧sinθhkl n=1 1  0.1542  sin   2  0.117  0.659 ∴ 2  2  sin 1   82.44 o

Chapter 3-70

3.17 Noncrystalline Solids

3.22

Chapter 3-71

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 non-directional (i.e., they are isotropic) in polycrystals with randomly oriented grains. Chapter 3-72