BNQ 30503 Chapter 1 (S).pdf

BNQ 30503 MATERIAL MA TERIAL ENGINE ENGINEERING ERING TECHNOLOGY CHAPTER 1: SOLID SOL ID STRUCTURE STRUCTURE By: Engr. Dr. Nasrul Fikry Che Pa http://bit.ly/19o6LoL

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Content’s Content’s Review

• Chapter 1: Solid Structure  – What you will learn?

• Crystal structures fundamental concepts, metallic crystal structures, crystallographic points, directions and planes, crystalline crystalline and noncrystalline noncrystalline materials, materials, point point defects, dislocations, importance of defects.

Content’s Content’s Review

• Chapter 1: Solid Structure  – What you will learn?

• Crystal structures fundamental concepts, metallic crystal structures, crystallographic points, directions and planes, crystalline crystalline and noncrystalline noncrystalline materials, materials, point point defects, dislocations, importance of defects.

Introduction Schematic representation of the most basic structural unit for quartz (all silicate material). Each atom of Si surrounded by 4 O2 atom whose center are located at the corner of a tetrahedron. Represented as 44−

Sketch of a unit cell for Quartz, composed of several interconnected 44− tetrahedral. 3

Introduction

Schematic diagram showing a large interconnected 44− tetrahedral. Observe the shape resembles quartz.

Photograph of 2 single crystals of quartz. 4

Fundamental Concepts Crystalline materials... •  Atoms pack in periodic, 3D arrays • Examples: metals, many ceramic & some polymers Non-crystalline materials... •  Atoms have no periodic packing • occurs for: - complex structures

crystalline SiO2 Si Oxygen

- rapid cooling - Lack systematic & regular arrangement of atoms over relatively large atomic distance

" Amorphous" = Noncrystalline -literally means ‘without form’

noncrystalline SiO2

“Atoms in crystalline solid are position in orderly & repeated patterns that are in contrast to the random & disordered atomic distribution found in

Metallic Crystal Structures • How can we stack metal atoms to minimize empty space?

2-dimensions

vs.

Imagine that these 2D layers are stacked to make 3D structures. Which one having a minimized empty spaces?

Metallic Crystal Structures General Description • Tend to be densely packed. • Assumptions in 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. - Electron cloud shields cores from each other • Have the simplest crystal structures. Next…. We

will examine four such structures...

Simple Cubic Structure (SC) • Rare due to low packing density (only Po, Polonium has this structure)

• Close-packed directions are cube edges. • Coordination # = 6 (# nearest neighbors)

Click once on image to start animation (Courtesy P.M. Anderson)

Atomic Packing Factor (APF)  APF =

Volume of atoms in unit cell* (V s) Volume of unit cell (V c)

*assume hard spheres • APF for a simple cubic structure = 0.52

atoms a

unit cell R=0.5a

 APF =

volume atom 4 p (0.5a) 3 1 3 a3

close-packed directions contains 8 x 1/8 = 1 atom/unit cell

volume unit cell

Body Centered Cubic Structure (BCC) • Atoms touch each other along cube diagonals. --Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.

ex: Cr, W, Fe ( ), Tantalum, Molybdenum

• Coordination # = 8

Click once on image to start animation (Courtesy P.M. Anderson)

2 atoms/unit cell: 1 center + 8 corners x 1/8

Atomic Packing Factor: BCC • APF for a body-centered cubic structure = 0.68 3a

a 2a

R

Close-packed directions: length = 4R = 3 a

a

atoms

4

2 unit cell 3  APF = volume unit cell

p ( 3a/4) 3 a3

volume atom

Face Centered Cubic Structure (FCC) • Atoms touch each other along face diagonals. --Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing. ex: Al, Cu, Au, Pb, Ni, Pt, Ag

• Coordination # = 12

Click once on image to start animation (Courtesy P.M. Anderson)

4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8

Atomic Packing Factor: FCC • APF for a face-centered cubic structure = 0.74 maximum achievable APF

Close-packed directions: length = 4R = 2 a 2a

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

a

atoms

4

4 unit cell 3  APF =

p ( 2 a/4) 3 a3

volume atom volume unit cell

FCC Stacking Sequence • ABCABC... Stacking Sequence • 2D Projection

B  A sites

 A

B sites

B

C B

C B

B C B

B

C sites

• FCC Unit Cell

 A B C

Hexagonal Close-Packed Structure (HCP) • ABAB... Stacking Sequence • 3D Projection

c

• 2D Projection

 A sites

Top layer 

B sites

Middle layer 

 A sites

Bottom layer 

a

• Coordination # = 12 • APF = 0.74 • c/a = 1.633 (long over short branch)

6 atoms/unit cell

ex: Cd, Mg, Ti, Zn

Exercise: 1. If the atomic radius of aluminum (FCC) is 0.143 nm, calculate the volume of its unit cell in cubic meters.

Theoretical Density, Density =

where

r =

Mass of  Atoms in Unit Cell

r =

n A VC N A

Total Volume of Unit Cell

n = number of atoms/unit cell  A = atomic weight VC = Volume of unit cell = a3 for cubic N A = Avogadro’s number  = 6.022 x 1023 atoms/mol

Periodic Table

Theoretical Density, • Ex: Cr (BCC)

R 

a 

atoms unit cell

r= volume unit cell

2 52.00 a3 6.022x1023

 A

= 52.00 g/mol

R

= 0.125 nm

n

= 2 atoms/unit cell

a = 4R/ 3 = 0.2887 nm

g mol

rtheoretical = 7.18 g/cm3 ractual = 7.19 g/cm3

atoms mol

Exercise: 1) Show that the APF for HCP is 0.74

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Exercise: 1) Zirconium has an HCP crystal structure and a density of 6.51 g/cm3.  – (a) What is the volume of its unit cell in cubic meters?  – (b) If the c/a ratio is 1.593, compute the values of c and a. Clues: =

n A VC N A

Exercise 2. Below are listed the atomic weight, density, and atomic radius for three hypothetical alloys. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination. A simple cubic unit cell is shown in Figure 3.24. Alloy

Atomic Weight (g/mol)

Density (g/cm3)

Atomic Radius (nm)

A

77.4

8.22

0.125

B

107.6

13.42

0.133

C

127.3

9.23

0.142

Fig. 3.24: Hard sphere unit cell representation of the simple cubic crystal structure.

Exercise 3. Rhenium has an HCP crystal structure, an atomic radius of 0.137 nm, and a c/a ratio of 1.615. Compute the volume of the unit cell for Re.

Densities of Material Classes In general rmetals > rceramics > rpolymers 30 Why? Metals have... • close-packing (metallic bonding) • often large atomic masses

Ceramics have... • less dense packing • often lighter elements

Polymers have...

Composites have... • intermediate values

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Platinum Gold, W Tantalum

10

Silver, Mo Cu,Ni Steels Tin, Zinc

   )

5   m4   c 3    /

   3

  g    (

• low packing density (often amorphous) • lighter elements (C,H,O)

Metals/  Alloys

2       r 1

0.5 0.4 0.3

Titanium  Aluminum Magnesium

Graphite/ Ceramics/ Semicond

Polymers

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

Crystals as Building Blocks • Some engineering applications require single crystals: -- diamond single crystals for abrasives

(Courtesy Martin Deakins, GE Superabrasives, Worthington, OH. Used with permission.)

• Properties of crystalline materials often related to crystal structure. -- Ex: Quartz fractures more easily along some crystal planes than others.

-- turbine blades

Polycrystals • Most engineering materials are polycrystals.

 Anisotropic  Adapted from Fig. K, color inset pages of Callister 5e. (Fig. K is courtesy of Paul E. Danielson, Teledyne Wah Chang  Albany)

1 mm

• Each "grain" is a single crystal. • If grains are randomly oriented, overall component properties are not directional. • Grain sizes typically range from 1 nm to 2 cm (i.e., from a few to millions of atomic layers).

Isotropic

Single crystal vs Polycrystals • Single Crystals

E (diagonal) = 273 GPa

-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.

E (edge) = 125 GPa

200 mm

Data from Table 3.3, Callister & Rethwisch 8e. (Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.)

 Adapted from Fig. 4.14(b), Callister & Rethwisch 8e . (Fig. 4.14(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].)

Polymorphism • Two or more distinct crystal structures for the same material (allotropy/polymorphism)

iron system titanium

, -Ti carbon

liquid BCC

1538ºC -Fe 1394ºC

diamond, graphite

FCC

-Fe

912ºC BCC

-Fe

Crystal Systems Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal. 7 crystal systems

14 Bravais crystal lattices

Fig. 3.4, Callister & Rethwisch 8e.

Three edge lengths a, b, and c and three interaxial angles α, β, and γ are the lattice constants

Exercise 4.

Below is a unit cell for a hypothetical metal. (a) To which crystal system does this unit cell belong? (b) What would this crystal structure be called?

(c) Calculate the density of the material, given that its atomic weight is 141 g/mol.

Point Coordinates z

•Point coordinates for unit cell center are ½½½ a/2, b/2, c/2

111

c

a

000

b

y

x The manner in which the q, r and s coordinates at point P within the unit cell are determined. The q coordinate (which is a fraction) corresponds to the distance qa along the x axis, where a is the unit cell edge length. The respective r and s coordinates for the y and z axes are determined similarly.

•Point coordinates for unit cell corner are 111

Exercise 5. List the point coordinates for all atoms that are associated with the FCC unit cell.

Crystallographic Directions z

 Algorithm 1. Vector repositioned (if necessary) to pass through origin. 2. Read off projections in terms of unit cell dimensions a, b, and c y 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas

[uvw]

x

ex: 1, 0, ½ => 2, 0, 1 => [201] -1, 1, 1 => [111]

where overbar represents a negative index

families of directions (crystallographically equivalent [spacing btw atoms is the same] e.g in cubic system)

Linear Density Linear Density of Atoms  LD =

[110]

No. of atoms centered on direction vector  Unit length of direction vector 

ex: linear density of Al in [110] direction a = 0.405 nm # atoms

2 = = 3.5 nm-1 LD

a

length

2a

Exercise 6. What are the indices for the directions indicated by the two vectors in the sketch below?

Exercise 7.

For tetragonal crystals, cite the indices of directions that are equivalent to each of the following directions:

(a) [001] (b) [110] (c) [010]

HCP Crystallographic Directions z

 Algorithm

a2

-

a3

1. Vector repositioned (if necessary) to pass through origin. 2. Read off projections in terms of unit cell dimensions a1, a2, a3, and c 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas [uvtw]

a2

a1  Adapted from Fig. 3.8(a), Callister & Rethwisch 8e.

-a3

a2

2

ex:

½, ½, -1, 0

=>

[1120]

a3

dashed red lines indicate projections onto a1 and a2 axes

a1 2

a1

HCP Crystallographic Directions • Hexagonal Crystals  – 4 parameter Miller-Bravais lattice coordinates are related to the direction indices (i.e., u'v 'w ') as follows. z

[ u 'v 'w ' ]  [ uvtw ] u

=

v

=

a2

-

a3 a1

Fig. 3.8(a), Callister & Rethwisch 8e.

1 3 1

(2 u ' - v ') (2 v ' - u ')

3 - (u +v )

= w = w' t

Exercise 8. Determine indices for the directions shown in the following hexagonal unit cells:

Crystallographic Planes

 Adapted from Fig. Fig. 3.10, Callister & Rethwisch 8e.

Crystallographic Planes • Miller Indices: Reciprocals Reciprocals of the (three) (three) axial axial intercepts intercepts for a plane, cleared cl eared of fractions & common multiples. All parallel parallel planes planes have have same Miller indices. • Algorithm 1. Read off inte intercep rcepts ts of plane with axes axes in terms of a, b, c 2. Take reciprocals of intercepts 3. Reduce Reduce to smallest smallest integer integer values values 4. Enclose Enclose in parentheses parentheses,, no commas i.e., (hkl )

Crystallographic Planes z example 1. Intercepts

3.

Reduction

a 1 1/1 1 1

4.

Miller Indices

(110) (110)

2.

Reciprocals

example 1. Intercepts 2. Reciprocals 3.

Reduction

a 1/2 1/½ 2 2

4.

Miller Indices

(100)

b 1 1/1 1 1

c

c

 1/ 0 0

b

a x

b

0 0

z

c

  1/ 1/

y

c

0 0

a x

b

y

Crystallographic Planes z

example 1. Intercepts 2. Reciprocals 3.

Reduction

4.

Miller Indices

a 1/2 1/½ 2 6

b 1 1/1 1 3

(634)

c c 3/4 1/¾ 4/3 4 a x







b

Family of Planes {hkl} Ex: {100} = (100), (010), (001), (100), (010), (001)

y

Exercise 9. Determine the Miller indices for the planes shown in the following unit cell:

10. Cite the indices of the direction that results from the intersection of each of the following pair of planes within a cubic crystal: (a) (100) and (010) planes, (b) (111) and (111) planes, and (c) (101) and (001) planes.

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Crystallographic Planes (HCP) • In hexagonal unit cells the same idea is used z

example 1. Intercepts 2. Reciprocals 3.

a1 1 1 1 1

Reduction

a2

 1/ 0 0

a3 -1 -1 -1 -1

c 1 1 1 1

a2

a3

4.

Miller-Bravais Indices

(1011)

a1 = h

a3 = i = -(h+k)

a2 = k

z=l

a1  Adapted from Fig. 3.8(b), Callister & Rethwisch 8e.

Planar Density •

Used to examine the atomic packing of crystallographic planes

•

E.g: Iron foil can be used as a catalyst. The atomic packing of the exposed planes is important. a)

Draw (100) crystallographic planes for Fe assuming that the structure is BCC.

b) Calculate the planar density.

Planar Density of (100) Iron • Solution: At T < 912ºC iron has the BCC structure. 2D repeat unit 4 3 = a R

(100)

3

Radius of iron R = 0.1241 nm atoms 2D repeat unit

Planar Density = area 2D repeat unit

1 a2

1

=

4 3 3

2 = 12.1

R

atoms atoms 19 = 1.2 x 10 nm2 m2

Exercise 11. (a) Derive planar density expressions for FCC (100) and (111) planes in terms of the atomic radius R.

(b) Compute the planar density values for these same two planes for nickel when R = 0.125 nm.

Summary from part 1 • Atoms may assemble into crystalline or amorphous structures. • Common metallic crystal structures are FCC, BCC, and HCP. Coordination number and atomic packing factor  are the same for both FCC and HCP crystal 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). • Crystallographic points, directions and planes are specified in terms of indexing schemes. Crystallographic directions and planes are related to atomic linear densities and planar densities.

Summary from part 1 (Cont.) • Materials can be single crystals or polycrystalline. Material properties generally vary with single crystal orientation (i.e., they are anisotropic), but are generally non-directional (i.e., they are isotropic) in polycrystals with randomly oriented grains. • Some materials can have more than one crystal is referred to as polymorphism (or allotropy).

structure. This

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Imperfections in Solids • Solidification- result of casting of molten material  – 2 steps

• Nuclei form • Nuclei grow to form crystals – grain structure • Start with a molten material – all liquid

nuclei

liquid

crystals growing

grain structure

• Crystals grow until they meet each other 

Polycrystalline Materials • Grain Boundaries • regions between crystals • transition from lattice of one region to that of the other • slightly disordered • low density in grain boundaries  – high mobility  – high diffusivity  – high chemical reactivity

Solidification •

Grains can be - equiaxed (roughly same size in all directions) - columnar (elongated grains) ~ 8 cm

heat flow Columnar in area with less undercooling

heat flow Shell of equiaxed grains due to rapid cooling (greater T) near wall

Grain Refiner - added to make smaller, more uniform, equiaxed grains.

Types of Imperfections • Vacancy atoms • Interstitial atoms • Substitutional atoms

Point defects

• Dislocations

Line defects

• Grain Boundaries

 Area defects

Point Defects in Metals • Vacancies: -vacant atomic sites in a structure.

Vacancy distortion of planes

• Self-Interstitials: -"extra" atoms positioned between atomic sites.

selfinterstitial distortion of planes

Equilibrium Concentration: Point Defects • Equilibrium concentration varies with temperature!

 Activation energy

No. of vacancies Total No. of atomic sites

 -Q v   = exp    kT   N

Nv

Temperature

Boltzmann's constant (1.38 x 10

-23 -5

J/atom-K)

(8.62 x 10 eV/atom-K) Each lattice site is a potential vacancy site

Estimating Vacancy Concentration • Find the equil. # of vacancies in 1 m 3 of Cu at 1000C. • Given: r = 8.4 g/cm3  A Cu = 63.5 g/mol Qv = 0.9 eV/atom N A = 6.02 x 1023 atoms/mol

0.9 eV/atom

 -Q   = exp  v  = 2.7 x 10-4  kT   N

Nv

For 1

m3 ,

N=

r x

N A  A Cu

1273 K 8.62 x 10-5 eV/atom-K x 1 m3 = 8.0 x 1028 sites

• Answer: Nv = (2.7 x 10-4)(8.0 x 1028) sites = 2.2 x 1025 vacancies

Exercise (Tutorial) 1. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy for vacancy formation of 0.55 eV/atom.

2. Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium number of vacancies at 500C (773 K) is 7.55  1023 m-3. The atomic weight and density (at 500C) for aluminum are, respectively, 26.98 g/mol and 2.62 g/cm3.

Imperfections in Metals Two outcomes if impurity (B) added to host (A): • Solid solution of B in A (i.e., random dist. of point defects)

OR Substitutional solid soln. (e.g., Cu in Ni)

Interstitial solid soln. (e.g., C in Fe)

• Solid solution of B in A plus particles of a new phase (usually for a larger amount of B) Second phase particle -- different composition -- often different structure.

Impurities in Solids • Specification of composition weight percent

C1

=

m1

x 100

m1  m2

m1 = mass of component 1

atom percent

'

C1

=

nm1

x 100

nm1  nm 2

nm1 = number of moles of component 1 ,

nm1 =

Derive an equation that relate C’1 & C1 ?  Ans: C’1 = [(C1 A2)/ [(C1 A2) + (C2 A1) ] ] x 100

  ()   ()

Additional Equation

Conversion of weight percent to atom percent (for two-element alloy)

′ = 2′ =

    +      + 

x 100 x 100 ′  2′ = 100    2  = 100

Conversion of atom percent to weight percent (for two-element alloy)

 = 2 =

Where,

 & 2 = weight percent   &  2 = Atomic weight

    +      + 

x 100 x 100

′  2′ = Atom percents

64

Additional Equation

Conversion of weight percent to Mass per unit volume (for 2 element alloy)



" = 2" =

3

X 10

  +  



X 103

   +  

" , = Concentration (kg/3 ) ρ , ρ2 = density in (g/3 )



ρ =  

Computation of density (for twoElement metal alloy)

ρ =

  =



  +       +  

  = Computation of atomic weight (for a two-element metal alloy)

 + 

     +  

  +    65

Line / Linear Defects Dislocations: • are line defects, • slip between crystal planes result when dislocations move, • produce permanent (plastic) deformation.

Schematic of Zinc (HCP): • before deformation

• after tensile elongation

slip steps

Imperfections in Solids • Linear Defects (Dislocations) Are one-dimensional defects around which atoms are misaligned

• Edge dislocation: extra half-plane of atoms inserted in a crystal structure

b perpendicular () to dislocation line

• Screw dislocation: spiral planar ramp resulting from shear deformation

b parallel () to dislocation line

Burger’s vector, b : measure of lattice distortion

Imperfections in Solids • Edge Dislocation

Motion of Edge Dislocation • Dislocation motion requires the successive bumping of a half plane of atoms (from left to right here). • Bonds across the slipping planes are broken and remade in succession.

Imperfections in Solids Screw Dislocation

b

Dislocation line

(b)

Burgers vector b (a)

Imperfections in Solids Mixed

Edge Screw

Imperfections in Solids • Dislocations are visible in electron micrographs

TEM micrograph of a titanium alloy in which the dark lines are dislocations. 51,450x.

TEM micrograph of dislocations in austenitic stainless steel

Dislocations & Crystal Structures • Structure: close-packed planes & directions are preferred.

view onto two close-packed planes.

close-packed directions close-packed plane (bottom)

• Specimens that were tensile tested.

close-packed plane (top)

Mg (HCP) tensile direction

 Al (FCC)

Significance of Dislocations 1. 2. 3.

It provide a mechanism for plastic deformation. It provide ductility in metals. We control the mechanical properties of a metal or alloy by interfering with the movement of dislocations. • Strain hardening • Solid solution strengthening • Grain size strengthening 4. It influence the electronic and optical properties of materials.

Interfacial Defects in Solid (Area) - Grain Boundaries • Boundary separating two small grains or crystals having different orientations. • Atoms are bonded less regularly along the grain boundary, result in boundary energy (similar to surface energy). • High angle boundaries have

higher energy. • Grain boundaries are

chemically reactive compared to grains because of the boundary energy. • Impurity atom often segregate

along grain boundaries because of the boundary energy.

Interfacial Defects in Solids • Phase Boundaries  – Exist in multiple phase materials.  – The different phases exist on each side of the boundary and it have its own relative physical and/or chemical characteristics.  – Important in determining Solder

the mechanical characteristic

Interface IMC

of multiphase metal alloys. Ni

Optical micrograph of intermetallic layer Cu between Nickel-gold surface finish and lead free solder.

500x

Interfacial Defects in Solids

• Twin boundary (plane)  – Essentially a reflection of atom positions across the twin plane.

(mirror-image)

• Stacking faults  – For FCC metals an error in ABCABC packing sequence  – Ex: ABCABABC