bda 10803 materials science

BDA 10803Materials ScienceSession 2021/2022Semester I

P M T S . D R . H A M I M A H B I N T I A B D. R A H M A ND e p t . o f M a n u f a c t u r i n g E n g i n e e r i n g

F K M P, U T H M

C h a p te r 2 :

M AT E R I A L S ST RU C T U R E

CHAPTER 2

har/jkp/fkmp/2021BDA 10803 2

2.1 Atomic structure and bonding

2.2 Crystal system and Bravais lattice

2.3 Metallic crystal structure

2.4 Crystallographic directions and planes

2.5 Crystal structure analysis (X-Ray Diffraction)

har/jkp/fkmp/2021BDA 10803 3

ISSUES TO ADDRESS...

• How do atoms assemble into solid structures?(for now, focus on metals)

• How does the density of a material depend onits structure?

• When do material properties vary with thesample (i.e., part) orientation?

har/jkp/fkmp/2021 BDA 10803 4

ATOMBasic Unit of an Element

Diameter : 10 –10 m.

Neutrally Charged

NucleusDiameter : 10 –14 m

Accounts for almost all mass

Positive Charge

Electron CloudMass : 9.109 x 10 –28 g

Charge : -1.602 x 10 –9 C

Accounts for all volume

ProtonMass : 1.673 x 10 –24 g

Charge : 1.602 x 10 –19 C

NeutronMass : 1.675 x 10 –24 g

Neutral Charge

STRUCTURE OF ATOMS

2.1 Atomic structure and bonding

2.1 ATOMIC AND BONDING STRUCTURE

Fundamental concepts → atomic number, atomic weight, atomic mass unit (amu), mole etc.

Electron in atoms → atomic models, quantum numbers, electron configurations, periodic table.

Primary interatomic bonds → ionic bonding, covalent bonding, metallic bonding.

Secondary bonding (Van Der Waals) → hydrogen bonding, polar-molecule-induced dipole bonds, permanent dipole bonds.

Nucleus: Z = # protons

orbital electrons: n = principal quantum number

n=3 2 1

= 1 for hydrogen to 94 for plutonium

N = # neutrons

Atomic mass A ≈ Z + N

Adapted from Fig. 2.1,

Callister 6e.

BOHR ATOM

• have discrete energy states

• tend to occupy lowest available energy state.

Electrons...

Callister 6e.

ELECTRON ENERGY STATES

• have complete s and p subshells

• tend to be unreactive.

Stable electron configurations...

Adapted from Table 2.2,

Callister 6e.

STABLE ELECTRON CONFIGURATIONS

• Why? Valence (outer) shell usually not filled completely.

• Most elements: Electron configuration not stable.

Electron configuration 1s1

1s2 (stable) 1s22s1 1s22s2 1s22s22p1 1s22s22p2 ...

1s22s22p6 (stable) 1s22s22p63s1 1s22s22p63s2 1s22s22p63s23p1 ...

1s22s22p63s23p6 (stable) ...

1s22s22p63s23p63d104s246 (stable)

Adapted from Table 2.2,

Callister 6e.

SURVEY OF ELEMENTS

• Columns: Similar Valence Structure

Electropositive elements:

Readily give up electrons

to become + ions.

Electronegative elements:

Readily acquire electrons

to become - ions.

Adapted from

Fig. 2.6,

Callister 6e.

THE PERIODIC TABLE

• Ranges from 0.7 to 4.0,

Smaller electronegativity Larger electronegativity

• Large values: tendency to acquire electrons.

Adapted from Fig. 2.7, Callister 6e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd

ELECTRONEGATIVITY

• Occurs between + and - ions.

• Requires electron transfer.

• Large difference in electronegativity required.

• Example: NaCl

IONIC BONDING

• Predominant bonding in Ceramics

Give up electrons Acquire electrons

Cl 3.0

Br 2.8

At 2.2

Li 1.0

Na 0.9

Rb 0.8

Cs 0.7

Fr 0.7

Be 1.5

Mg 1.2

Ca 1.0

Sr 1.0

Ba 0.9

Ra 0.9

Ti 1.5

Cr 1.6

Fe 1.8

Ni 1.8

Zn 1.8

As 2.0

Adapted from Fig. 2.7, Callister 6e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd

EXAMPLES: IONIC BONDING

• Requires shared electrons

• Example: CH4

C: has 4 valence e,

needs 4 more

H: has 1 valence e,

needs 1 more

Electronegativities

are comparable.

Adapted from Fig. 2.10, Callister 6e.

COVALENT BONDING

• Molecules with nonmetals

• Molecules with metals and nonmetals

• Elemental solids (RHS of Periodic Table)

• Compound solids (about column IVA)

Cl 3.0

Br 2.8

At 2.2

Li 1.0

Na 0.9

Rb 0.8

Cs 0.7

Fr 0.7

Be 1.5

Mg 1.2

Ca 1.0

Sr 1.0

Ba 0.9

Ra 0.9

Ti 1.5

Cr 1.6

Fe 1.8

Ni 1.8

Zn 1.8

As 2.0

C(diamond)

Si 1.8

Ga 1.6

Ge 1.8

Sn 1.8

Pb 1.8

EXAMPLES: COVALENT BONDING

• Arises from a sea of donated valence electrons

(1, 2, or 3 from each atom).

• Primary bond for metals and their alloys

Adapted from Fig. 2.11, Callister 6e.

METALLIC BONDING

Arises from interaction between dipoles

• Permanent dipoles-molecule induced

• Fluctuating dipoles

-general case:

-ex: liquid HCl

-ex: polymer

Callister 6e.

Adapted from

Fig. 2.14,

Callister 6e.

Adapted from

Fig. 2.14,

Callister 6e.

SECONDARY BONDING

Covalent

Metallic

Secondary

Bond Energy

Large!

Variable

large-Diamond

small-Bismuth

Variable

large-Tungsten

small-Mercury

smallest

Comments

Nondirectional (ceramics)

Directional

semiconductors, ceramics

polymer chains)

Nondirectional (metals)

Directional

inter-chain (polymer)

inter-molecular

SUMMARY: BONDING

• Bond length, r

• Bond energy, Eo

• Melting Temperature, Tm

Tm is larger if Eo is larger.

PROPERTIES FROM BONDING: TM

• Elastic modulus, E

• E ~ curvature at ro

Ao = E

Elastic modulus

larger Elastic Modulus

smaller Elastic Modulus

Energy

ro unstretched length

E is larger if Eo is larger.

PROPERTIES FROM BONDING: E

• Coefficient of thermal expansion, a

• a ~ symmetry at ro

a is larger if Eo is smaller.

= a (T2-T1) L

coeff. thermal expansion

PROPERTIES FROM BONDING: a

Ceramics

(Ionic & covalent bonding):

Metals

(Metallic bonding):

Polymers

(Covalent & Secondary):

Large bond energy

large Tm

large Esmall a

Variable bond energy

moderate Tm

moderate Emoderate a

Directional Properties

Secondary bonding dominates

small T

small Elarge a

SUMMARY: PRIMARY BONDS

2.2 Crystal system and Bravais lattice

2.2 CRSYTAL SYSTEMS & BRAVAIS LATTICE SYSTEM

Crystal – a solid composed of atoms, ions or molecules arranged in a pattern that is repeated in three dimensions.

Solid materials – classified according to the regularity with which atoms or ions are arranged.

Atomic arrangement – determine the solid materials microstructure and properties.

Examples : ductility and strength

MATERIALS

- Atoms are positioned/situated in an orderly and repeated pattern, 3D arrays

- Examples : all metals, many ceramic and some polymers

- Atoms are distributed randomly and disordered.

-Occurs for– complex structure

- rapid cooling

- Examples : polymer & a few ceramics

MATERIALS AND PACKING

• have the simplest 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.

METALLIC CRYSTALS…

CRYSTAL SYSTEM

Crystal structure – a regular three-dimensional pattern of atoms or ions in space.

Space lattice – a three-dimensional array of points coinciding with atom positions (or sphere centers).

Lattice point – atom position in the crystal lattice

Unit cell – small groups of atoms form a repetitive pattern

Crystal lattice

▪ 7 crystal systems of unit cell for metallic crystal structures;

1. Ciubic

2. Tetragonal

3. Orthorhombic

4. Rhombohedral

5. Hexagonal

6. Monoclinic

7. Triclinic

▪ These systems can be divided into 14 Bravais Unit Cell System

▪Space group = Lattice identifier + known symmetry relationships

▪Molecules within the crystal will most likely pack with symmetry

14 Bravais Unit Cell Crsytal System

2.3 Metallic crystal structure

2.3 METALLIC CRYSTAL STRUCTURE

Most of the metallic materials - 3 types of crystal structure :

Eg: γ – Fe, Cu, Al, Ag, Eg: α – Fe, Cr, W Eg: Mg, Ti, Zn, Cd

Atomic packing factor (APF);

Vtotal sphere = natom Vatom

Vunit cell Vunit cell

Coordination number;

number of nearest-neighbor or touching atoms for each

atom in the crystal structure

* assume hard spheres

• Rare due to poor packing (only Polonium, Po has this structure)

SIMPLE CUBIC (SC) STRUCTURE

52 % of the unit cell is occupied with atoms.

ATOMIC PACKING FACTOR SC Structure

BODY-CENTERED CUBIC (BCC) STRUCTURE

34Ra =

a = unit cell length

R = atomic radius

68% of the unit cell is occupied with atoms.

FACE-CENTERED CUBIC (FCC) STRUCTURE

a = unit cell length

R = atomic radius

74% of the unit cell is occupied with atoms.

HEXAGONAL CLOSED PACKED (HCP) STRUCTURE

ATOMIC PACKING FACTOR HCP Structure

THEORETICAL DENSITY,

Example :

Copper has the FCC crystal structure with atomic radius, R = 0.1278 nm. Assume the atoms to be hard spheres and packed as close together as possible along the FCC unit cell cross-section.

Calculate the theoretical volume density of copper in g/cm3.

(Atomic mass of Cu = 63.54 g/mol)

SOLUTION :

For FCC,

nm 0.361 2

nm) 4(0.1278

373 cm) 10(0.361a −=

3-23 cm 104.70=

23323-Ac

g/cm 8.89atom/mol)10 (6.022 cm 10 x 4.70

g/mol 63.54 x 4

nAρ =

Volume of unit cell, V =

2.4 Crystallographic directions and planes

2.4 CRYSTALLOGRAPHIC PLANES AND DIRECTIONS

Deformation in metal such as forging, drawing etc. –moves according to certain planes and directions inthe crystal structure.

Example : Ferum (Fe) – Magnetic effect is strong in[100] direction compare to [111] direction.

Miller Indices – used to explain the planes anddirections position in crystal.

Atomic position in BCC unit cellAtomic position in cubic unit cell

CRYSTALLOGRAPHIC DIRECTIONS

• A line or vector between 2 (two) points in crystal structure.

• Steps in determination of the three directional indices :

1. A vector of convenient length is positioned such that it passes through theorigin of the coordinate system. Any vector may be translated throughoutthe crystal lattice without alteration, if parallelism is maintained.

2. The length of the vector PROJECTION on each of the three axes isdetermined; they are measured in terms of the unit cell dimensions a, band c.

3. These three numbers are multiplied or divided by a common factor toreduce them to the smallest INTEGER values.

4. The three direction indices are enclosed by square brackets with noseparating commas, thus : [ ].

→ +ve

→ -ve

−−−

All parallel direction vectors have the same direction indices.

Crystallographically equivalent directions – the atom spacing along each direction is the same.

other directions of a family <110> and <111>

- INDICES OF A FAMILY

]1[000],1[000],1[[001],[010],[100],100 →

har/jkp/fkmp/2021 BDA 10803

Examples of Crystal Directions

(1 0 1)

Notes :

= ( _ _ _ )

= ( 0 _ _ )

= ( _ 0 _ )

= ( _ _ 0 )

Problems

Draw the following crystallograhic planes in cubic unit cells :

(234) (d)

(221) (c)

(010) (b)

(101) (a)

An important relationships for the cubic system (only)→ the direction indices of a direction perpendicular to a crystal plane

are the same as the Miller indices of that plane.Example : the [100] direction is perpendicular to the (100) crystal

plane.

• PLANES OF A FAMILY – sets of equivalent lattice planes are related by thesymmetry of the crystal system (SAME ATOMIC ARRANGEMENTS within allthose planes)

Determine the planes indices of the cubic direction shown in figures below.

(a) (b) (c)

(d) (e) (f)

CRYTALLOGRAPHIC PLANES IN HEXAGONAL UNIT CELLS

The four coordinate axes (a1, a2, a3

and c) of he HCP crystal structure unit cell

➢ The HCP crystal planes indices, are donated by the letters h, k, iand l.

➢ Enclosed in parentheses as (hkil)

➢ a1, a2 and a3 – basal axes which make 120 with each other.

➢ c axis – the vertical axis located at the center of the unit cell.

PLANAR ATOMIC DENSITY (P)

planeof Area

plane selected by dintersecte are centers whose atomsof no. Equiv.ρ

Example Problem

Calculate the planar atomic density p on the plane of the a iron BCC lattice inatoms per square millimeter. The lattice constant, a of a iron is 0.287 nm.

• The units of planar density = m-2, nm-2

(b) Areas of atoms in BCC unit cell cut by the (110) plane

(a) A BCC atomic-site unit cell showing a shaded (110) pane

Solution :

• Equivalent number of atoms intersected by the (110) plane ;

1 atom at the center + (4 x ¼ atoms at four corners of plane) = 2 atoms

atoms/mm 17.2x10nm17.2atoms/Pρ

nm) (0.287 2

atoms 2Pρ

Area of the plane (110) ;

(√2 a)(a) = √2 a2

The planar atomic density is

LINEAR ATOMIC DENSITY (l)

vector directionof Length

vector direction on centered atomsof Numberρ

Example Problem

Calculate the linear atomic density l in the [110] direction in the copper crystal lattice in atoms per millimeter. Copper is FCC and has a lattice constant of 0.361 nm.

• The units of linear density = m-1 , nm-1

Diagram for calculating the atomic linear density in the [110] direction in an FCC unit cell

Solution :

The number of atomic diameters intersected by this length of line are

½ + 1 + ½ = 2 atoms.

atoms/mm 10 x 3.92atoms/nm 3.92lρ

nm) (0.361 2

atoms 2lρ

Length of the direction vector = √2 a

The linear atomic density is

2.5 Crystal structure analysis (X-Ray

Diffraction)

2.5 Crystal Structure Analysis (X-Ray Diffraction)

Spacing between planes → distance between 2 parallel planes having similar miller indices, given by (hkl is miller indices for plane).

The analysis is done by using the X-Ray Diffractometer Machine.

XRD Profile

Carbon monoxidemolecules arrangedon a platinum (111)surface.

Iron atoms arrangedon a copper (111)surface. These Kanjicharacters representthe word “atom”.

Atoms can be arranged and imaged!!!

SCANNING TUNNELING MICROSCOPY

Polymorph materials – example: Iron (Fe)

SUMMARY

Atoms may assemble into crystalline or amorphousstructures.

We can predict the density of a material, provided we knowthe atomic weight, atomic radius, and crystal structure(e.g., FCC, BCC, HCP).

Material properties generally vary with single crystalorientation (i.e., they are anisotropic), but properties aregenerally non-directional (i.e., they are isotropic) inpolycrystals with randomly oriented grains.

bda 10803 materials science

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