The metallurgical fundamental

STRUCTURE OF METALS • What happens when many atoms come together to form a solid?

• Regular structures (crystalline) or Irregular structures

(amorphous)

Crystalline:

“A solid characterized by periodic or repeating arrays over large atomic

distances”

“long-range order” , “repetitive 3D pattern”

Often opaque

Most metals

Amorphous (Non-crystalline)

A solid which may have short-range order, but no long-range order.

“without form”

Often transparent

Ceramic glasses

Amorphous metals

Some polymers are completely amorphous

Quartz: crystalline SiO2 Glass: amorphous SiO2

What is the difference between quartz and glass?

Long-range order Only short-range order

Crystal Structures

• Describes the way atoms (ions) arranged in 3D or “lattice” space in terms

of unit cell geometry & atom position within a unit cell.

• Allotropes: >1 crystal structure

Unit Cell: - basic building block (simplest structural unit) of crystal structure

- indicates crystal structure geometry & atom position

repeats

repeats Many unit cells form a crystalline solid

• FCC, BCC, and HCP crystal structures

Crystal Systems

• Scheme by which crystal structures are classified only by unit cell geometry

(not atom position).

• 7 possible unit cell geometries

- Ex: cubic, hexagonal

• 7 possible crystal systems

• See next Table and Fig.

• Unit cell geometries determined by 6 lattice parameters

1. 3 edge lengths (a, b, c)

2. 3 interaxial angles (a, b, )

FCC crystal structure

BCC crystal structure

HCP crystal structure

all edges = all angles =

CRYSTAL SYSTEMS

BRAVAIS LATTICES

• Many of the seven crystal systems have

variations of the basic unit cell.

• A.J. Bravais showed that 14 standard unit

cell could describe all possible lattice

networks. These Bravais lattices are

illustrated in the next Figure

The 14 Bravais conventional unit cells grouped

according to crystal system.

Types of Models to Describe Crystal Structures

Hard Sphere Model (a): - “big atoms”

- Atoms are like spheres

- Atoms touch neighbor

- Radius of hard sphere = atomic radius

- Atomic radius: distance between two

nuclei of two touching atoms

r = d/2 d

Reduced Sphere Model (b)

-Center of atoms represented as small

circles

Aggregate of many atoms (c)

Main Crystal Structures for Metals: FCC, BCC, HCP.

Face-centered cubic crystal structure (FCC)

1 2

3 4

1, 2, 3, 4:

4 corner nearest atoms

Consider

this

atom

Body-centered cubic crystal structure (BCC)

Hexagonal close-packed crystal structure (HCP)

The coordination number (#)

CN = the number of closest neighbors to which an atom is bonded.

CN=6 in cubic structure; CN=12 in FCC; CN =8 in BCC.

Number of atoms for unit cell, n=4 for FCC n=2 for

BCC

•8 corner atoms shared by 8 cells : 8 x 1/8 = 1

•6 face atoms shared by 2 cells : 6 x ½ = 3

•Center atom (1) shared by no other cells = 1

Face-centered cubic crystal structure (FCC)

1 2

3 4

1, 2, 3, 4:

4 corner nearest atoms

Consider

this

atom

The coordination

CN =12

• APF for a simple cubic structure = 0.52

APF =

a 3

4

3 p (0.5a) 3 1

atoms

unit cell atom

volume

unit cell

volume

ATOMIC PACKING FACTOR (APF)

APF = Volume of atoms in unit cell*

Volume of unit cell

*assumed hard spheres

close-packed directions

a

R=0.5a

aR

• APF for a body-centered cubic structure = p3/8 = 0.68

Unit cell contains: 1 + 8 x 1/8 = 2 atoms/unit cell

Adapted from

Fig. 3.2,

Callister 6e.

ATOMIC PACKING FACTOR: BCC

Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell

a

• APF for a body-centered cubic structure = p/(32) = 0.74

(best possible packing of identical spheres)

Adapted from

Fig. 3.1(a),

Callister 6e.

ATOMIC PACKING FACTOR: FCC

HEXAGONAL CLOSE-PACKED

STRUCTURE (HCP) • Unit cell – hexagonal

• 6 atoms form regular hexagon, surrounding one atom in center. Another plane is situated halfway up unit cell (c-axis), with 3 additional atoms situated at interstices of hexagonal (close-packed) planes.

• Metals with this crystal strcs. = Cd, Mg, Zn, Ti, Co,etc.

Fig. For the HCP crystal structures (a) a reduced-sphere unit cell (a and c represent the

short and long edge lengths, respectively), (b) an aggregate of many atoms.

• Unit cell has two lattice parameters a and c. Ideal ratio c/a = 1.633

• The coordination #,CN =12 (same as in FCC)

• Number of atoms per unit cell, n=6.

– 3 mid-plane atoms shared by no other cells = 3

– 12 hexagonal corner atoms shared by 6 cells: 12 x 1/6 = 2

– 2 top/bottom plane center atoms shared by 2 cells: 2 x ½ = 1

• Atomic packing factor, APF = 0.74 (same as in FCC close-packed structure)

HCP

CLOSE-PACKED CRYSTAL

STRUCTURES (FCC AND HCP)

• Both FCC and HCP crystal structures have APF of 0.74 (maximum possible value)

• Both FCC and HCP crystal structures may be generated by the stacking of close-packed planes – Close-packed planes = planes having a max. atom

or sphere-packing density)

• The difference btw. the 2 strcs. is in the stacking sequence.

Portion of a close-packed

plane of “A atoms”

AB stacking sequence for

close-packed atomic planes

Triangular vertix “up” = “B”

Triangular vertix “down” = “C”

*At this point (2 layers), AB or AC stacking sequences are equivalent

BOTH FCC and HCP are CLOSE-PACKED

Form 2 types of triangular

vertices “B” and “C”

A plane of close-packed

“B atoms” is added on top of the

close-packed plane of “A

atoms” over the “B vertices”

HCP: ABAB… Stacking Sequence

(ACACAC would be equivalent)

Centers of 3rd layer of “A atoms”

are positioned directly above

the “A atoms” of the 1st layer

1

2

3

c

a

A sites

B sites

A sites

• 2D Projection (looking down)

Bottom layer (1 or A)

Middle layer (2 or B)

Top layer (3 or A)

FCC: ABCABC… Stacking Sequence

Centers of 3rd close-packed

layer of “C atoms” are

positioned directly above the

“C vertices” of 1st layer

1

2

3

A sites

B B

B

B B

B B

C sites

C C

C A

B

B sites B B

B

B B

B B

B sites C C

C A

C C

C A

A B

C

• 2D Projection (looking down)

FCC unit cell

BCC

•All atoms are identical; the center atom

is shaded differently only for ease of viewing

•Atoms touch each other along cube (body) diagonal

FCC

•All atoms are identical; corner atoms are shaded

differently only for ease of viewing

•Atoms touch each other along face diagonal

Bottom layer

Middle layer

Top layer

HCP

Models

DENSITY

where n = number of atoms/unit cell

A = atomic weight

VC = Volume of unit cell = a3 for cubic

NA = Avogadro’s number

= 6.023 x 1023 atoms/mol

Density =

VC NA

n A =

Cell Unit of Volume Total

Cell Unit in Atoms of Mass

Densities of Material Classes

metals > ceramics > polymers

Why?

Data from Table B1, Callister 7e.

(g

/cm

)

3

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibers

Polymers

1

2

2 0

30 B ased 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

G raphite

Silicon

Glass - soda Concrete

Si nitride Diamond Al oxide

Zirconia

H DPE, PS PP, LDPE

PC

PTFE

PET PVC Silicone

Wood

AFRE *

CFRE *

GFRE*

Glass fibers

Carbon fibers

A ramid 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

Polymorphism in Alloys

Austenitic stainless: FCC

form stabilized with Ni

Ti-6Al-4V: BCC form

stabilized with aluminum

CRYSTALLOGRAPHIC POINTS,

DIRECTIONS & PLANES

• In crystalline materials, often necessary to specify points, directions and planes within unit cell and in crystal lattice

• Three numbers (or indices) used to designate points, directions (lines) or planes, based on basic geometric notions

• The three indices are determined by placing the origin at one of the corners of the unit cell, and the coordinate axes along the unit cell edges

Unit Cells and Unit Cell Vectors

a

b

c

Lattice parameters

axial lengths: a, b, c

interaxial angles: a, b,

unit vectors:

In general: a ≠ b ≠ c

a ≠ b ≠

a

b

c

All period unit cells may be described

by these vectors and angles.

31

CRYSTALLOGRAPHIC

POINTS • The position of any

point located within a unit cell may be specified in terms of its coordinates as fractional multiples of the unit cell edge lengths.

• With considering point P, point P

x y z

a b c

q r s

EXAMPLE: POINT COORDINATES

• Locate the point (1/4 1 ½)

• Specify point coordinates for all atom

positions for a BCC unit cell

– Answer: 0 0 0, 1 0 0, 1 1 0, 0 1 0, ½ ½ ½,

0 0 1, 1 0 1, 1 1 1, 0 1 1

CRYSTALLOGRAPHIC

DIRECTIONS • Defined as a line between 2 points: a vector

• Steps for finding the 3 indices denoting a direction: – Determine the point positions of a beginning point (X1 Y1 Z1)

and a ending point (X2 Y2 Z2) for direction, in terms of unit cell edges

– Calculate difference between ending and starting point

– Multiply the differences by a common constant to convert them to the smallest possible integers u, v, w

– The three indices are not separated by commas and are enclosed in square brackets: [uvw]

– If any of the indices is negative, a bar is placed in top of that index

EXAMPLE

3. Enclosure [120]

1. Projections 1/2 1 0

2. Reduction 1 2 0

x y z

EXAMPLE

CRYSTALLOGRAPHIC

PLANES • Crystallographic planes specified by 3 Miller indices as (hkl)

• Procedure for determining h,k and l:

– If plane not passes through origin, translate plane or choose new origin

– Determine intercepts of planes on each of the axes in terms of unit cell edge lengths (lattice parameters). Note: if plane has no intercept to an axis (i.e., it is parallel to that axis intercept is infinity).

– example: intercepts are (½ ¼ ½)

– Determine reciprocal of the three intercepts (2 4 2)

– If necessary, multiply these three numbers by a common factor which converts all the reciprocals to small integers (1 2 1)

– The three indices are not separated by commas and are enclosed in curved brackets: (h k l) (121)

– If any of the indices is negative, a bar is placed in top of that index

EXAMPLES z

x

y a b

c

4. Miller Indices (110)

a b c z

x

y a b

c

4. Miller Indices (100)

1. Intercepts 1 1

2. Reciprocals 1/1 1/1 1/

1 1 0 3. Reduction 1 1 0

1. Intercepts 1/2

2. Reciprocals 1/½ 1/ 1/

2 0 0 3. Reduction 2 0 0

a b c

EXAMPLE z

x

y a b

c

4. Miller Indices (634)

1. Intercepts 1/2 1 3/4 a b c

2. Reciprocals 1/½ 1/1 1/¾

2 1 4/3

3. Reduction 6 3 4

THREE IMPORTANT CRYSTAL

PLANES

( 1 0 0) (1 1 1)(1 1 0)

Close-packed planes = planes having a max. atom or sphere-packing

density) These are important during deformation and “slip”; planes tend to

slip or slide along planes with high density and along directions with high

density.

LINEAR & PLANAR DENSITIES

Crystal Structures of Selected

Pure Metals

• Face Centered Cubic (FCC) – Aluminum (Al)

– Copper (Cu)

– Nickel (Ni)

• Close packed hexagonal (CPH) – Titanium

– Magnesium

• Body Centered Cubic (BCC) – Iron (Fe)

– Chromium (Cr)

– Tungsten (W)

CRYSTALLINE MATERIALS

• SINGLE CRYSTALS

– Crystals can be single crystals where the whole solid

is one crystal. Then it has a regular geometric

structure with flat faces.

• POLYCRYSTALLINE

– Many small crystals or grains. The grains have

different crystallographic orientation. There exist

atomic mismatch within the regions where grains

meet. These regions are called grain boundaries.

• Single Crystals

-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.

200 mm

Data from Table 3.3,

Callister 6e.

(Source of data is R.W.

Hertzberg, Deformation

and Fracture Mechanics

of Engineering

Materials, 3rd ed., John

Wiley and Sons, 1989.)

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].)

SINGLE VS POLYCRYSTALS

*

•Point defects: vacancy or missing atoms, interstitial atom or extra atom in the lattice or impurity foreign atom that has replaced the atom of pure metal

•Linear defections: dislocations

•Planar imperfections: grain boundaries and phase boundaries

•Volume or bulk imperfections: voids, inclusions, other phases, cracks

Defects

Point defects

Solute atoms can dissolve in a solid solvent

metal in two ways:

– Interstitial Solution

• Solvent atoms fit within interstitial voids of the

solvent crystal

• Examples: C in iron; O in titanium

– Substitutional Solution

• Solvent atoms replace atoms on solvent lattice

• Examples: Ni in iron; Cu in aluminum

Solid Solutions

Schematic illustration of types of defects in single crystal

lattice: interstitial, vacancy, substitutional.

Interstitial voids

Planar imperfections -Grain Boundaries

• Separate regions of different crystal orientation

• Low bond density (energy ~ 0.5 J/m2)

• Open Structure (fast diffusion/impurity segregation)

• Influence mechanical properties – Low TIncrease strength & ductility

– High T Decrease strength

• Grain boundaries – the surfaces that separate individual

grains

• Rapid cooling – smaller grains

• Slow cooling – larger grains

• Grain size- at room temperature a large grain size is generally

associated with low strength, low hardness, and low ductility

(ductility is a solid material's ability to deform under tensile

stress)

• Grain size is measured by counting the number of grains in a

given area or by counting the number of grains that intersect

a length of line randomly drawn on an enlarged photograph

of the grains

GRAIN SIZE

Line defects Dislocations

line-defects in the orderly arrangement of a

metal’s atomic structure.

Because a slip plane containing a dislocation

requires less shear stress to allow slip than does

a plane in a perfect lattice, dislocations are the

most significant defects that explain the

discrepancy between the actual and theoretical

strengths of metals.

*

*

Theoretical Shear Strength and Tensile

Strength

• Theoretical shear stress is the shear stress to cause permanent deformation in a perfect crystal.

• Theoretical or ideal tensile strength of material is the tensile stress required to break the atomic bonds between two neighboring atomic planes.

• The actual strength of metals is approximately one to two orders of magnitude lower than the theoretical strengths. The discrepancy can be explained in terms of imperfections in the crystal structure.

30/~10/2

max GGbetweena

bG

p

10/max E

*

•Dislocations can become entangled and interfere with each other and be impeded by barriers such as grain boundaries, impurities, and inclusions in the material. The increased shear stress required to overcome entanglements and impediments results in an increase in overall strength and hardness of the metal and is known as work hardening or strain hardening. (Ex. Cold rolling, forging, drawing)

• The dislocation motions result in the macroscopic plastic deformation (tensile test); the macroscopic sample is composed of many grains and in each grain ( about 100 mm), there are many dislocations (0.01 mm) moving on slip plane and in slip direction.

Slip Systems

• Deformation (dislocation) occurs on preferential

crystallographic planes and directions, called slip

systems.

• The slip plane/direction is the plane/direction with the

most closely packed atoms.

6x2=12 4x3=12 1x3=3

Tensile test

Stress- strain curve

Plastic strain

Propagation of a defect