crystal structure lec

7/31/2019 Crystal Structure Lec

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Crystal Structure



SUBCONTENT:

Crystal structures. Fundamental Concept

Unit cell

Material crystal system

Efficiency of atomic packing

Polymorphism and allotropic transformation

Crystallography plane and direction



You should be able:

• Differentiate between crystalline, noncrystalline, single

crystal and polycrsytal.

• Derive the relationship between unit cell edge length andatomic radius for face centered and body centered cubic

• Compute density and atomic packing factors for crystal

• Specify the miller indices for crystallography plane anddirection

• Relate the crystal structure with material properties

LEARNING OBJECTIVE



Fundamental concept:

Classification of solid materials

Solid materials

CrystallineMaterial

Single Crystal polycrystal

Noncrsytallinematerial

(Amorphous)



Fundamental concept: Crystalline

material

Crystalline material- atoms, molecules or ion packed in aregularly ordered, repeating pattern, extending in 3 spatial

dimension.

Properties of crystalline materials depend on the crystalstructure

Crystal structure

is a unique arrangement of atoms in a crystal.

composed of a unit cell



• Formation crystal structure depends

– Chemistry of fluid

– Condition under which is being solidified

– Ambient pressure

• Crystallization-Process forming crystalline

structures.



• atoms pack in periodic, 3D arrays

• typical of:

Crystalline materials...

-metals-many ceramics-some polymers

• atoms have no periodic packing • occurs for:

Noncrystalline materials...

-complex structures

-rapid cooling

crystalline SiO2

noncrystalline SiO2

"Amorphous" = Noncrystalline



•No recognizable long-

range order

•Completely ordered

•In segments

•Entire solid is made up

of atoms in an orderlyarray

Amorphous

Polycrystalline

Crystal

•Atoms are disordered

•No lattice

•All atoms arranged ona common lattice

•Different latticeorientation for eachgrain



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• Single Crystals

-Properties vary withdirection: anisotropic.

-Example: the modulusof elasticity (E) in BCC iron:

• Polycrystals

-Properties may/may notvary with direction.

-If grains are randomly

oriented: isotropic.(Epoly iron = 210 GPa)

-If grains are textured,anisotropic.

200 mm

SINGLE VS POLYCRYSTALS



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• Some engineering applications require single crystals:

--diamond singlecrystals for abrasives

--turbine blades

• Most engineering materials are polycrystals.

grain

http://sites.google.com/site/polymorphismmyhomepage/investigating-material-failures/Fracture_Eng006.jpg?attredirects=0

http://sites.google.com/site/polymorphismmyhomepage/investigating-material-failures/Fracture_Eng006.jpg?attredirects=0



• Unit cell - smallest structural unit or building block that can describe thecrystal structure. Repetition of the unit cell generates the entire crystal.

• Primitive unit cell- smallest possible unit cell one can construct.

• Lattice parameters- spacing between unit cells in various direction.

http://sites.google.com/site/danaashkenazi1/crystals006.jpg?attredirects=0



Concept test

which one isunit cell



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The atomic arrangements inthe different crystal systems

affect mechanical behavior.

1.5 CRYSTAL SYSTEM AND CRYSTALLOGRAPHY

•By adding additional lattice point to 7 basic crystal systems – form 14 Bravais

lattice.

•However, most metals exhibit one of three crystal structure – BCC, FCC & HCP

•The crystal systems are a grouping of crystal structures that used to describetheir lattice.

•Each crystal system consists of a set of three axes in a particular geometricalarrangement.

•There are only 7 crystal systems that atoms can pack together to produce aninfinite 3D space lattice :•cubic, hexagonal, tetragonal, rhombodhedral, orthorhombic, monoclinic, triclinic.

Crystal System





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BRAVAIS LATTICE

Figure : The 14 Bravais conventional unitcells are grouped according to crystalsystem. The dots indicate lattice points that

when allocated on faces or at corners, areshared by other identical lattice unit cells.

BRAVAIS LATTICE - describe the geometric arrangement of the lattice points andthe translational symmetry of the crystal.



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FACE CENTRED CUBIC(FCC)

HEXAGONAL CLOSEDPACK (HCP)

SIMPLE CUBIC (SC)

BODY CENTRED CUBICBCC

CRYSTAL STRUCTURE OF METALS



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Types of crystal structure Examples

Simple cubic Cube where atoms lie on a grid.

No. of atom at corner = 8x1/8 = 1 atom

Total no. of atom in 1unit cell = 1 atom

Manganese

Body centered cubic(BCC)

*Lower ductility but a higher yield strength than FCC metals.

Cube with an atom at each corner and one in the center.

No. of atom at corner = 8x1/8 = 1 atomNo. of atom at center = 1 atomTotal no. of atom in 1unit cell = 2 atoms

Chromium,Tungsten(W),Molybdenum,Vanadium(V)

CRYSTAL STRUCTURE OF METALS

Most metals (about 90%) crystallize upon solidification into three densely packed crystal

structures as shown below (BCC,FCC,HCP). However, simple cubic crystals are rather rare whichcan be viewed as simple cubic grid.



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Types of crystal structure Examples

Face centered cubic (FCC)

* Atoms for FCC are more densely packed than BCC.

Cube with an atom at each corner, one in the center and one in thecenter of each side of the cube.

No. of atom at corner = 8x1/8 = 1 atom

No. of atom at face = 6x1/2 = 3 atomTotal no. of atom in 1unit cell = 4 atoms

Aluminum,Nickel, Copper,Gold, Lead,Platinum

Hexagonal close-packed(HCP)

•Because of the spacing of the lattice structure, rows of atoms do not easily slide over one another in HCP --- have lower plasticity and ductility than cubic structures.

*Atoms are as densely packed as in the FCC unit cell. Difference--- Arrangement of the atoms

Hexagon with a reference atom that is surrounded by 12 atoms that areequal distance from the reference atom.

No. of atom at corner for top face = 6x1/6 = 1 atomNo. of atom at corner for bottom face = 6x1/6 = 1 atomNo. of atom at centre = 3 atomsNo. of atom at centre for top face = 1x1/2 = 1/2 atom

No. of atom at centre for bottom face = 1x1/2 = 1/2 atomTotal no. of atom in 1unit cell = 6 atoms

Magnesium,Beryllium, Zinc,Cadmium,Titanium,Zirconium (Zr)



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a (lattice constant) Atoms/unitcell

Coordi-nation No.

Examples

Simplecubic

a = 2R 1 6 CsCl

BCCa = 4R/√3

2 8 Many metals:α-Fe, Cr, Mo, W

FCCa = 4R/√2

or

a = 2R√2

4 12 Many metals :Ag, Au, Cu, Pt

HCP a = 2Rc = 1.633 a

6 12 Many metals :Co, Mg, Ti, Zn

Summary



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APF = no. of atom/unit cell, n x volume of atoms in the unit cell, (Vs)volume of the unit cell, (Vc)

ATOMIC PACKING FACTOR

•Atomic packing factor (APF) is defined as the efficiency of atomic arrangementin a unit cell.

•It is used to determine the most dense arrangement of atoms. It is because howthe atoms are arranged determines the properties of the particular crystal.

•It is represented mathematically by :



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EXAMPLE

• APF for a simple cubic structure = 0.52

Calculate the APF for Simple Cubic (SC)?



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EXERCISE

a) BCC b) FCC

Calculate the APF for BCC and FCC ?



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DENSITY COMPUTATIONS

• A knowledge of the crystal structure of a metallicsolid permits computation of its density through the

relationship :

Where

ρ = n AVc NA

n = number of atoms associated with each unit cell

A = atomic weightVc = volume of the unit cellNA = Avogadro’s number (6.023 x 1023 atoms/mol)



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Calculate the density for nickel (simple cubic structure).Note that the unit cell edge length (a) for nickel is 0.3524 nm.

EXAMPLE

The volume (V) of the unit cell is equal to the cell-edge length (a) cubed.

V = a3 = (0.3524 nm)3 = 0.04376 nm3

Since there are 109 nm in a meter and 100 cm in a meter, there must be 107 nm in a cm.109 x 1m = 107 nm/cm1 m 100 cm

We can therefore convert the volume of the unit cell to cm3 as follows.4.376 x 10-2 nm3 x [1 cm ]3 = 4.376 x 10-23 cm 3

107 nm

The mass of a nickel atom can be calculated from the atomic weight of this metal and Avogadro’snumber.

58.69g Ni x 1 mol = 9.746 x 10-23 g/atom 1(9.746 x 10-23 g/unit cell) = 2.23 g/cm3 1 mol 6.023 x 1023 atoms 4.376 x 10-23 cm 3 /unit cell



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Crystalstructure, 20oC

Density of solid,20oC (g/cm3)

Copper has an atomic radius of 0.128 nm, FCC crystal structure and an atomicweight of 63.5 g/mol. Compute its density and compare the answer with itsmeasured density.

Element Symbol

AtomicNumber

Atomicweight(amu)

Atomicradius(nm)

Chromium Cr 24 52.00 7.19 BCC 0.125

Cobalt Co 27 58.93 8.9 HCP 0.125

Copper Cu 29 63.55 8.94 FCC 0.128

Solution :

ρ = n A

Vc NA

EXERCISE



Polymorphism allotropic

transformation

• the ability of a solid material to exist in more

than one form or crystal structure.

• Example: Carbon and iron



Allotropy: Carbon

Graphite Diamond



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ALLOTROPIC TRANSFORMATION

• A material that can exist in more than one lattice structure (depending ontemperature-heating@cooling) allotropic.

• An allotropic material is able to exist in two or more forms having variousproperties without change in chemical composition.

• E.g : When metals are heated slowly from room temperature to their meltingpoints (liquid state), most metals (Cu,Al) do not change their crystalline latticestructure before becoming a liquid. However, this in not the case with steel, Ti,iron and many iron alloys.

Upon heating pure iron experiences two changes in crystal structure



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912°C

1394°C

1538°C

768°C

Allotropy of iron(Fe)

BCC (delta iron-high T) FCC (austenite-moderate T) BCC (alfa iron-low T)

Upon heating, pure iron experiences two changes in crystal structure.

At room temperature, it exists as ferrite,or α iron (BCC).

When we heat it to 912°C, it experiences an allotropic transformation to austenite,or γ iron (FCC).

At 1394°C, austenite reverts back to a BCC phase called δ ferrite.



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Miller indices is used to label the planes and directions of atoms in a crystal.

Why Miller indices is important?To determine the shapes of single crystals, the interpretation of X-raydiffraction patterns and the movement of a dislocation , which may determinethe mechanical properties of the material.

MILLER INDICES



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i) Determine the points at which a given crystal planeintersects the three axes, say at (a,0,0),(0,b,0), and (0,0,c). Ifthe plane is parallel an axis, it is given an intersection ∞.

ii) Take the reciprocals of the three integers found in step (i).

iii) Label the plane (hkl). These three numbers are expressedas the smallest integers and negative quantities are indicatedwith an overbar,e.g : a.

MILLER INDICES OF A PLANE

How to determine crystal plane indices?

Figure : Planes with different Millerindices in cubic crystals

Axis X Y ZInterceptions

Reciprocals

Reduction (if necessary)

Enclosed (h k l )



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Axis X Y Z

Interceptions ½ 1 ½

Reciprocals 2 1 2


Enclosed (2 1 2)

EXAMPLE : CRYSTAL PLANE INDICES

0

1

1

1

10

1

1

½

½

Axis X Y Z

Interceptions 1 1 1

Reciprocals


Enclosed (1 1 1)



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Axis X Y Z

Interceptions 1 1 ∞

Reciprocals 1 1 0


Enclosed (1 1 0)

0

1

1

1

Axis X Y Z

Interceptions 1 ∞ ½

Reciprocals 1 0 2


Enclosed (1 0 2)

5/6

⅓

Two numbers in one axes

0



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Axis X Y Z

Interceptions ∞ 1 ½

Reciprocals 0 1 2


Enclosed (0 1 2)

½

Plane pass through origin

0

Axis X Y Z

Interceptions ∞ 1 ∞

Reciprocals 0 1 0


Enclosed (0 1 0)0



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Determine the Miller Indices plane

for the following figure below?

Answer :a) (121)

b) (210)

c) (111)

a)

0

1

1

1

Axis x y z

Intercepts

Reciprocals

Reduction(if necessary)

Enclosed ( )

½



Answer :a) (121)

b) (210)

c) (111)

b)

0

1

1

Axis x y z

Intercepts

Reciprocals


Enclosed ( )

½



Answer :

a) (121)

b) (210)

c) (111)c)

0

1

1

1

c)

Axis x y z

Intercepts

Reciprocals

Reduction

(if necessary)

Enclosed ( )



Axis x y z

Intercepts

Reciprocals


Enclosed ( )

Answer :a) (121)

b) (210)

c) (111) c)

0

1

1

1

d)

⅗ ⅕



Determine the Miller indices of a cubic crystal plane that intersects the position coordinates

A (1, 1/4, 0), B (1, 1, 1/2) ,C (3/4, 1, 1/4) and D (1/2, 1, 0) ?

Answer : ( 6 4 6 )

Axis x y z

Intercepts

Reciprocals

Reduction

(if necessary)

Enclosed ( )



Draw the following Miller Indices

plane.

a) ( 1 0 0 )

b) ( 0 0 1 )

c) ( 1 0 1 )

d) ( 1 1 0 )

e) ( 0 1 1 )

f) ( 2 2 1 )

g) ( 1 3 1 )

h) ( 0 1 0 )



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MILLER INDICES OF A DIRECTION

How to determine crystal direction indices?i) Determine the length of the vector

projection on each of the three axes,based on .

ii) These three numbers are expressed as thesmallest integers and negative quantitiesare indicated with an overbar.

iii) Label the direction [hkl]. Figure : Examples of direction

Axis X Y Z

Head (H) x2 y2 z2

Tail (T) x1 y1 z1

Head (H) –Tail (T) x2-x1 y2-y1 z2-z1


Enclosed [h k l]

* No reciprocal involved.



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Axis X Y Z

Head (H) 1 0 ½

Tail (T) 0 0 0

Projection (H-T) 1 0 ½

Reduction (if necessary) * 2 2 0 1

Enclosed [2 0 1]

EXAMPLE : CRYSTAL DIRECTION INDICES

0½

1

1

Axis X Y Z

Head (H) 1 0 0

Tail (T) 0 1 0

Projection (H-T) 1 1 0

Enclosed [1 1 0]0

1

1

1

1



Determine the Miller Indices direction

for the following figure below? a)

Answer :

a) (121)

b) (210)

c) (111)c)

0

1

1

1

Axis x y z

Head

Tail

Projection(Head – Tail)


Enclosed [ ]



Answer :

a) (121)

b) (210)

c) (111)

b)

0

1

1

1

½

½

Axis x y z

Head

Tail



Enclosed [ ]



Answer :a) (121)

b) (210)

c) (111) c)

0

1

1

1c) Axis x y z

Head

Tail

Projection

(Head–

Tail)


Enclosed [ ]



Answer :a) (121)

b) (210)

c) (111) c)

0

1

1

1d) Axis x y z

Head

Tail

Projection

(Head–

Tail)


Enclosed [ ]



Answer :a) (121)

b) (210)

c) (111) c)

0

1

1

1e) Axis x y z

Head

Tail

Projection

(Head–

Tail)


Enclosed [ ]

½



Answer :a) (121)

b) (210)

c) (111) c)

0

1

1

1f) Axis x y z

Head

Tail

Projection

(Head–

Tail)


Enclosed [ ]

½



Determine the direction indices of the cubicdirection between the position coordinates

TAIL (3/4, 0, 1/4) and HEAD (1/4, 1/2, 1/2)?

Axis x y z

Head

Tail



Enclosed [ ]

Answer : [ 2 2 1 ]



Determine the direction indices of the cubicdirection between the position coordinates

TAIL (1, 0, 1/2) and HEAD (0, 1, 1/3)?

Axis x y z

Head

Tail



Enclosed [ ]

Answer : [ 6 6 1 ]



Draw the following Miller Indices

direction.

a) [ 1 0 0 ]

b) [ 1 1 1 ]

c) [ 1 1 0 ]

d) [ 1 1 0 ]

e) [ 1 1 2 ]



NOTE (for plane and direction):

• PLANE

Make sure you enclosed your final answer in brackets (…) with noseparating commas → (hkl)

• DIRECTION Make sure you enclosed your final answer in brackets (…) with no

separating commas → [hkl]

• FOR BOTH PLANE AND DIRECTION

Negative number should be written as follows :

-1 (WRONG)

1 (CORRECT)

Final answer for labeling the plane and direction should not have fraction number do a reduction.

crystal structure lec

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