the science and engineering of materials -...

12/22/2008١Chapter 3 -

The Science and Engineering of Materials

W. CallisterR. Askeland

P. Phulé

S. M. RabieePhD in Biomedical Materials Engineering

&

Assistant Professor in Department Of Mechanical Engineering

Babol (Noshirvani) University Of Technology

٢Chapter 3 -

o What is Materials Science and Engineering?o Classification of Materialso Functional Classification of Materialso Classification of Materials Based on Structureo Environmental and Other Effectso Materials Design and Selection

Introduction to Materials Science and Engineering

٣Chapter 3 -

What is Materials Science and Engineering?

o Materials Science and Engineeringo Composition means the chemical make-up of a

material.o Structure means a description of the arrangements

of atoms or ions in a material.o Synthesis is the process by which materials are

made from naturally occurring or other chemicals.o Processing means different ways for shaping

materials into useful components or changing their properties.

٤Chapter 3 -

Classification of Materials

q Metals and Alloysq Ceramics, Glasses,and Glass-ceramicsq Polymers (plastics), Thermoplastics and Thermosetsq Composite Materials

٥Chapter 3 -

Types of Materials• Metals:

– Strong, ductile– high thermal & electrical conductivity– opaque, reflective.

• Polymers: Covalent bonding à sharing of e’s– Soft, ductile, low strength, low density– thermal & electrical insulators– Optically translucent or transparent.

• Ceramics: ionic bonding (refractory) –compounds of metallic & non-metallic elements (oxides, carbides, nitrides, sulfides)– Brittle, glassy, elastic– non-conducting (insulators)

٦Chapter 3 -

©2003 B

rooks/Cole Publishing / Thom

son Learning™

Figure : Representative strengths of various categories of materials

٧Chapter 3 -

Functional Classification of Materials

q Aerospaceq Biomedicalq Electronic Materialsq Energy Technology and Environmental Technologyq Magnetic Materialsq Photonic or Optical Materialsq Smart Materialsq Structural Materials

٨Chapter 3 -

©2003 B


son Learning™

Figure: Functional classification of materials. Notice that metals, plastics, and ceramics occur in different categories. A limited number of examples in each category is provided

٩Chapter 3 -

Example of Biomedical Material – Hip Implant

Adapted from Fig. 22.26, Callister 7e.

١٠Chapter 3 -

Hip Implant• Key problems to overcome

– fixation agent to hold acetabular cup

– cup lubrication material– femoral stem – fixing agent

(“glue”)– must avoid any debris in cup

Adapted from chapter-opening photograph, Chapter 22, Callister 7e.

Femoral Stem

Ball

AcetabularCup and Liner

١١Chapter 3 -

Classification of Materials-Based on Structure

q Crystalline material is a material comprised of one or many crystals. In each crystal, atoms or ions show a long-range periodic arrangement.q Single crystal is a crystalline material that is made of only one crystal (there are no grain boundaries).q Grains are the crystals in a polycrystalline material.q Polycrystalline material is a material comprised of many crystals (as opposed to a single-crystal material that has only one crystal).q Grain boundaries are regions between grains of a polycrystalline material.

١٢Chapter 3 -

Environmental and Other Effects

Effects of following factors must be accounted for in design to ensure that components do not fail unexpectedly:

q Temperature q Corrosionq Fatigueq Strain Rate

١٣Chapter 3 -

Materials Design and Selection

q Density is mass per unit volume of a material, usually expressed in units of g/cm3 or lb/in.3q Strength-to-weight ratio is the strength of a material divided by its density; materials with a high strength-to-weight ratio are strong but lightweight.

١٤Chapter 3 -

• Use the right material for the job.

• Understand the relation between properties,structure, and processing.

• Recognize new design opportunities offeredby materials selection.

Course Goals:SUMMARY

١٥Chapter 3 -

ISSUES TO ADDRESS...

• What promotes bonding?

• What types of bonds are there?

• What properties are inferred from bonding?

BONDING AND PROPERTIES

١٦Chapter 3 -

Atomic Structure • atom – electrons – 9.11 x 10-31 kg

protonsneutrons

• atomic number = # of protons in nucleus of atom= # of electrons of neutral species

• A [=] atomic mass unit = amuAtomic wt = wt of 6.023 x 1023 molecules or atoms

1 amu/atom = 1g/mol

C 12.011H 1.008 etc.

} 1.67 x 10-27 kg

١٧Chapter 3 -

Calculate the number of atoms in 100 g of silver.

Example SOLUTION

The number of silver atoms is =)868.107(

)10023.6)(100( 23

molg

molatomsg ×

=5.58 × 1023

Example :Calculate the Number of Atoms in Silver

١٨Chapter 3 -

Atomic Structure

• Valence electrons determine all of the following properties

1) Chemical2) Electrical 3) Thermal4) Optical

١٩Chapter 3 -

Electronic Structure• Electrons have wavelike and particulate

properties. – This means that electrons are in orbitals defined by a

probability.– Each orbital at discrete energy level determined by

quantum numbers.

Quantum # Designationn = principal (energy level-shell) K, L, M, N, O (1, 2, 3, etc.)l = subsidiary (orbitals) s, p, d, f (0, 1, 2, 3,…, n-1)ml = magnetic 1, 3, 5, 7 (-l to +l)

ms = spin ½, -½

٢٠Chapter 3 -

©2003 B


son Learning™

Figure:The atomic structure of sodium, atomic number 11, showing the electrons in the K, L, and M quantum shells

٢١Chapter 3 -

©2003 B


son Learning™

Figure : The complete set of quantum numbers for each of the 11 electrons in sodium

٢٢Chapter 3 -

• Most elements: Electron configuration not stable.SURVEY OF ELEMENTS

Electron configuration

(stable)

...

... 1s 22s22p 63s23p 6 (stable)... 1s 22s22p 63s23p 63d 10 4s24p 6 (stable)

Atomic #

18...36

Element1s 11Hydrogen1s 22Helium1s 22s13Lithium1s 22s24Beryllium1s 22s22p 15Boron1s 22s22p 26Carbon

...1s 22s22p 6 (stable)10Neon1s 22s22p 63s111Sodium1s 22s22p 63s212Magnesium1s 22s22p 63s23p 113Aluminum

...Argon...Krypton

٢٣Chapter 3 -

Electron Configurations• Valence electrons – those in unfilled shells• Filled shells more stable• Valence electrons are most available for

bonding and tend to control the chemical properties

– example: C (atomic number = 6)

1s2 2s2 2p2

valence electrons

٢٤Chapter 3 -

Ionic bond – metal + nonmetal

donates acceptselectrons electrons

Dissimilar electronegativities

ex: MgO Mg 1s2 2s2 2p6 3s2 O 1s2 2s2 2p4

[Ne] 3s2

Mg2+ 1s2 2s2 2p6 O2- 1s2 2s2 2p6

[Ne] [Ne]

٢٥Chapter 3 -

• Occurs between + and - ions.• Requires electron transfer.• Large difference in electronegativity required.• Example: NaCl

Ionic Bonding

Na (metal) unstable

Cl (nonmetal) unstable

electron

+ -CoulombicAttraction

Na (cation) stable

Cl (anion) stable

٢٦Chapter 3 -

Ionic Bonding• Energy – minimum energy most stable

– Energy balance of attractive and repulsive terms

Attractive energy EA

Net energy EN

Repulsive energy ER

Interatomic separation r

Adapted from Fig. 2.8(b), Callister 7e.

EN = EA + ER

٢٧Chapter 3 -

• Predominant bonding in Ceramics

Adapted from Fig. 2.7, Callister 7e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by Cornell University.

Examples: Ionic Bonding

Give up electrons Acquire electrons

NaClMgOCaF2CsCl

٢٨Chapter 3 -

C: has 4 valence e-,needs 4 more

H: has 1 valence e-,needs 1 more

Electronegativitiesare comparable.


Covalent Bonding• similar electronegativity ∴ share electrons• bonds determined by valence – s & p orbitals

dominate bonding• Example: CH4

shared electrons from carbon atom

shared electrons from hydrogen atoms

H

H

H

H

C

CH4

٢٩Chapter 3 -

Primary Bonding• Metallic Bond -- delocalized as electron cloud

• Ionic-Covalent Mixed Bonding

% ionic character =

where XA & XB are Pauling electronegativities

%)100(x

1− e− (X A−XB )2

4

ionic 70.2% (100%) x e1 characterionic % 4)3.15.3( 2

=

−=

−−

Ex: MgO XMg = 1.3XO = 3.5

٣٠Chapter 3 -

Arises from interaction between dipoles

• Permanent dipoles-molecule induced

• Fluctuating dipoles

-general case:

-ex: liquid HCl


Adapted from Fig. 2.14,Callister 7e.

SECONDARY BONDING

asymmetric electronclouds

+ - + -secondary bonding

HH HH

H2 H2

secondary bonding

ex: liquid H2

H Cl H Clsecondary bonding

secondary bonding+ - + -

٣١Chapter 3 -

TypeIonic

Covalent

Metallic

Bond EnergyLarge!

Variablelarge-Diamondsmall-Bismuth

Variablelarge-Tungstensmall-Mercury

CommentsNondirectional (ceramics)

Directional(semiconductors, ceramicspolymer chains)

Nondirectional (metals)

Summary: Bonding

٣٢Chapter 3 -

Ceramics(Ionic & covalent bonding):

Metals(Metallic bonding):

Polymers(Covalent & Secondary):

Large bond energylarge Tmlarge Esmall α

Variable bond energymoderate Tmmoderate Emoderate α

Directional PropertiesSecondary bonding dominates

small Tmsmall Elarge α

Summary: Primary Bonds

٣٣Chapter 3 -


• 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?

The Structure of Crystalline Solids

٣٤Chapter 3 -

• atoms pack in periodic, 3D arraysCrystalline materials...

-metals-many ceramics-some polymers

• atoms have no periodic packingNoncrystalline materials...

-complex structures-rapid cooling

crystalline SiO2

noncrystalline SiO2"Amorphous" = NoncrystallineAdapted from Fig. 3.22(b),Callister 7e.

Adapted from Fig. 3.22(a),Callister 7e.

Materials and Packing

Si Oxygen

• typical of:

• occurs for:

٣٥Chapter 3 -

Crystal Systems

7 crystal systems

14 crystal lattices

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

a, b, and c are the lattice constants

٣٦Chapter 3 -

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

empty space?2-dimensions

vs.

Now stack these 2-D layers to make 3-D structures

٣٧Chapter 3 -

• Tend to be densely packed.• Reasons for dense packing:

- Typically, only one element is present, so all atomicradii 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.

We will examine three such structures...

Metallic Crystal Structures

٣٨Chapter 3 -

• Rare due to low packing denisty (only Po has this structure)• Close-packed directions are cube edges.

• Coordination # = 6(# nearest neighbors)

(Courtesy P.M. Anderson)

Simple Cubic Structure (SC)

٣٩Chapter 3 -

• APF for a simple cubic structure = 0.52

APF = a3

43

π (0.5a) 31atoms

unit cellatom

volume

unit cellvolume

Atomic Packing Factor (APF)

APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres


close-packed directions

a

R=0.5a

contains 8 x 1/8 = 1 atom/unit cell

٤٠Chapter 3 -

• Coordination # = 8



• Atoms touch each other along cube diagonals.--Note: All atoms are identical; the center atom is shaded

differently only for ease of viewing.

Body Centered Cubic Structure (BCC)

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

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

٤١Chapter 3 -

Atomic Packing Factor: BCC

a

APF =

43

π ( 3a/4)32atoms

unit cell atomvolume

a3unit cellvolume

length = 4R =Close-packed directions:

3 a

• APF for a body-centered cubic structure = 0.68

aR

Adapted from Fig. 3.2(a), Callister 7e.

a2

a3

٤٢Chapter 3 -




• Atoms touch each other along face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded

differently only for ease of viewing.

Face Centered Cubic Structure (FCC)

ex: Al, Cu, Au, Pb, Ni, Pt, Ag

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

٤٣Chapter 3 -

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

maximum achievable APF

APF =

43

π ( 2a/4)34atoms

unit cell atomvolume

a3unit cellvolume

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

Unit cell contains:6 x1/2 + 8 x1/8

= 4 atoms/unit cella

2 a

Adapted fromFig. 3.1(a),Callister 7e.

٤٤Chapter 3 -

A sites

B B

B

BB

B B

C sites

C C

CAB

B sites

• ABCABC... Stacking Sequence• 2D Projection

• FCC Unit Cell

FCC Stacking Sequence

B B

B

BB

B B

B sitesC C

CAC C

CA

AB

C

٤٥Chapter 3 -


• ABAB... Stacking Sequence

• APF = 0.74

• 3D Projection • 2D Projection

Adapted from Fig. 3.3(a),Callister 7e.

Hexagonal Close-Packed Structure (HCP)

6 atoms/unit cell

ex: Cd, Mg, Ti, Zn

• c/a = 1.633

c

a

A sites

B sites

A sites Bottom layer

Middle layer

Top layer

٤٦Chapter 3 -

Theoretical Density, ρ

where n = number of atoms/unit cellA = atomic weight VC = Volume of unit cell = a3 for cubicNA = Avogadro’s number

= 6.023 x 1023 atoms/mol

Density = ρ =

VCNA

n Aρ =

CellUnitofVolumeTotalCellUnitinAtomsofMass

٤٧Chapter 3 -

• Ex: Cr (BCC) A = 52.00 g/molR = 0.125 nmn = 2

ρtheoretical

a = 4R/ 3 = 0.2887 nm

ρactual

aR

ρ = a3

52.002atoms

unit cell molg

unit cellvolume atoms

mol

6.023 x1023

Theoretical Density, ρ

= 7.18 g/cm3

= 7.19 g/cm3

٤٨Chapter 3 -

Densities of Material Classesρmetals > ρceramics > ρpolymers

Why?

Data from Table B1, Callister 7e.

ρ(g

/cm

)3

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibersPolymers

1

2

20

30Based on data in Table B1, Callister

*GFRE, CFRE, & AFRE are Glass,Carbon, & Aramid Fiber-ReinforcedEpoxy composites (values based on60% volume fraction of aligned fibers

in an epoxy matrix).10

345

0.30.40.5

Magnesium

Aluminum

Steels

Titanium

Cu,NiTin, Zinc

Silver, Mo

TantalumGold, WPlatinum

GraphiteSiliconGlass -sodaConcrete

Si nitrideDiamondAl oxide

Zirconia

HDPE, PSPP, LDPEPC

PTFE

PETPVCSilicone

Wood

AFRE*CFRE*GFRE*Glass fibers

Carbon fibersAramid 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

٤٩Chapter 3 -

Figure: Coordinates of selected points in the unit cell. The number refers to the distance from the origin in terms of lattice parameters.

٥٠Chapter 3 -

Crystallographic Directions

1. Vector repositioned (if necessary) to pass through origin.

2. Read off projections in terms of unit cell dimensions a, b, and c

3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvw]

ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]

-1, 1, 1

families of directions <uvw>

z

x

Algorithm

where overbar represents a negative index

[ 111 ]=>

y

٥١Chapter 3 -

Determine the Miller indices of directions A, B, and Cin Figure

Example: Determining Miller Indices of Directions

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

Figure . Crystallographic directions and coordinates

٥٢Chapter 3 -


Figure . Equivalency of crystallographic directions of a form in cubic systems.

٥٣Chapter 3 -

٥٤Chapter 3 -

ex: linear density of Al in [110] direction

a = 0.405 nm

Linear Density• Linear Density of Atoms ≡ LD =

a

[110]

Unit length of direction vectorNumber of atoms

# atoms

length

13.5 nma2

2LD −==

٥٥Chapter 3 -

HCP Crystallographic Directions

1. Vector repositioned (if necessary) to pass through origin.

2. Read off projections in terms of unitcell dimensions a1, a2, a3, or c

3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvtw]

[ 1120 ]ex: ½, ½, -1, 0 =>


dashed red lines indicate projections onto a1 and a2 axes a1

a2

a3

-a32

a2

2a1

-a3

a1

a2

zAlgorithm

٥٦Chapter 3 -

HCP Crystallographic Directions• Hexagonal Crystals

– 4 parameter Miller-Bravais lattice coordinates are related to the direction indices (i.e., u'v'w') as follows.

=

=

=

'wwt

v

u

)vu( +-

)'u'v2(31 -

)'v'u2(31 -=

]uvtw[]'w'v'u[ →

Fig. 3.8(a), Callister 7e.

-a3

a1

a2

z

٥٧Chapter 3 -

Crystallographic Planes• Miller Indices: Reciprocals of the (three) axial

intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices.

• Algorithm1. Read off intercepts of plane with axes in

terms of a, b, c2. Take reciprocals of intercepts3. Reduce to smallest integer values4. Enclose in parentheses, no

commas i.e., (hkl)

٥٨Chapter 3 -

Determine the Miller indices of planes A, B, and C in Figure

Example: Determining Miller Indices of Planes


Figure . Crystallographic planes and intercepts

٥٩Chapter 3 -

SOLUTIONPlane A1. x = 1, y = 1, z = 12.1/x = 1, 1/y = 1,1 /z = 13. No fractions to clear4. (111)Plane B

1. The plane never intercepts the z axis, so x = 1, y = 2, and z = 2.1/x = 1, 1/y =1/2, 1/z = 03. Clear fractions:1/x = 2, 1/y = 1, 1/z = 04. (210)Plane C

1. We must move the origin, since the plane passes through 0, 0, 0. Let’s move the origin one lattice parameter in the y-direction. Then, x = , y = -1, and z =2.1/x = 0, 1/y = 1, 1/z = 03. No fractions to clear.

)010( .4

∞

∞∞

٦٠Chapter 3 -

Crystallographic Planesz

x

ya b

c

4. Miller Indices (110)

example a b cz

x

ya b

c


1. Intercepts 1 1 ∞2. Reciprocals 1/1 1/1 1/∞

1 1 03. Reduction 1 1 0

1. Intercepts 1/2 ∞ ∞2. Reciprocals 1/½ 1/∞ 1/∞

2 0 03. Reduction 2 0 0

example a b c

٦١Chapter 3 -

Crystallographic Planesz

x

ya b

c•

••


example1. Intercepts 1/2 1 3/4

a b c

2. Reciprocals 1/½ 1/1 1/¾2 1 4/3

3. Reduction 6 3 4

(001)(010),

Family of Planes {hkl}

(100), (010),(001),Ex: {100} = (100),

٦٢Chapter 3 -

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

example a1 a2 a3 c

4. Miller-Bravais Indices (1011)

1. Intercepts 1 ∞ -1 12. Reciprocals 1 1/∞

1 0 -1-1

11

3. Reduction 1 0 -1 1

a2

a3

a1

z

Adapted from Callister 7e.

٦٣Chapter 3 -

Example:Drawing Direction and Plane

Draw (a) the direction and (b) the plane in a cubic unit cell.

1]2[1 10]2[


Figure Construction of a (a) direction and (b) plane within a unit cell

٦٤Chapter 3 -

SOLUTION

a. Because we know that we will need to move in the negative y-direction, let’s locate the origin at 0, +1, 0. The ‘‘tail’’ of the direction will be located at this new origin. A second point on the direction can be determined by moving +1 in the x-direction, 2 in the y-direction, and +1 in the z direction [Figure 3.24(a)].

b. To draw in the plane, first take reciprocals of the indices to obtain the intercepts, that is:

x = 1/-2 = -1/2 y = 1/1 = 1 z = 1/0 =

Since the x-intercept is in a negative direction, and we wish to draw the plane within the unit cell, let’s move the origin +1 in the x-direction to 1, 0, 0. Then we can locate the x-intercept at 1/2 and the y-intercept at +1. The plane will be parallel to the z-axis [Figure 3.24(b)].

10]2[

∞

٦٥Chapter 3 -

Determine the Miller-Bravais indices for planes A and B and directions C and D in Figure

Example: Determining the Miller-Bravais Indices for

Planes and Directions


Figure 3.25 Miller-Bravais indices are obtained for crystallographic planes in HCP unit cells by using a four-axis coordinate system. The planes labeled A and B and the direction labeled C and D are those discussed in Example 3.11.

٦٦Chapter 3 -

Example SOLUTION

Plane A1. a1 = a2 = a3 = , c = 12. 1/a1 = 1/a2 = 1/a3 = 0, 1/c = 13. No fractions to clear4. (0001)

Plane B1. a1 = 1, a2 = 1, a3 = -1/2, c = 12. 1/a1 = 1, 1/a2 = 1, 1/a3 = -2, 1/c = 13. No fractions to clear4.

Direction C1. Two points are 0, 0, 1 and 1, 0, 0.2. 0, 0, 1, -1, 0, 0 = 1, 0, 13. No fractions to clear or integers to reduce.4.

∞

)1211(

113]2[or ]011[

٦٧Chapter 3 -

SOLUTION (Continued)

Direction D1. Two points are 0, 1, 0 and 1, 0, 0.2. 0, 1, 0, -1, 0, 0 = -1, 1, 03. No fractions to clear or integers to reduce.4. 100]1[or ]101[

٦٨Chapter 3 -

٦٩Chapter 3 -

Planar Density of (100) IronSolution: At T < 912°C iron has the BCC structure.

(100)

Radius of iron R = 0.1241 nm

R3

34a =

Adapted from Fig. 3.2(c), Callister 7e.

2D repeat unit

= Planar Density =a2

1atoms

2D repeat unit

= nm2

atoms12.1m2

atoms= 1.2 x 10191

2

R3

34area2D repeat unit

٧٠Chapter 3 -

o Interstitial sites - Locations between the ‘‘normal’’ atoms or ions in a crystal into which another - usually different - atom or ion is placed. Typically, the size of this interstitial location is smaller than the atom or ion that is to be introduced.

o Cubic site - An interstitial position that has a coordination number of eight. An atom or ion in the cubic site touches eight other atoms or ions.

o Octahedral site - An interstitial position that has a coordination number of six. An atom or ion in the octahedral site touches six other atoms or ions.

o Tetrahedral site - An interstitial position that has a coordination number of four. An atom or ion in the tetrahedral site touches four other atoms or ions.

Interstitial Sites

٧١Chapter 3 -


Figure . The location of the interstitial sites in cubic unit cells. Only representative sites are shown.

٧٢Chapter 3 -

Example Calculating Octahedral Sites

Calculate the number of octahedral sites that uniquelybelong to one FCC unit cell.

SOLUTION

The octahedral sites include the 12 edges of the unit cell, with the coordinates

211,,0

211,,1

210,,1

21,0,0

,121,0 ,1

21,1 ,0

21,1 ,0

21,0

1,1,21 0,1,

21 1,0,

21 0,0,

21

plus the center position, 1/2, 1/2, 1/2.

٧٣Chapter 3 -

Which sites will cations occupy?Site Selection

1. Size of sites– does the cation fit in the site

2. Stoichiometry– if all of one type of site is full the

remainder have to go into other types of sites.

3. Bond Hybridization

٧٤Chapter 3 -

٧٥Chapter 3 -

• Coordination # increases with

Coordination # and Ionic Radii

Adapted from Table 12.2, Callister 7e.

2

rcationranion

Coord#

< 0.155

0.155 - 0.225

0.225 - 0.414

0.414 - 0.732

0.732 - 1.0

3

4

6

8

linear

triangular

TD

OH

cubic




ZnS(zincblende)

NaCl(sodium chloride)

CsCl(cesium chloride)

rcationranion

Issue: How many anions can youarrange around a cation?

٧٦Chapter 3 -

Cation Site Size• Determine minimum rcation/ranion for OH site

(C.N. = 6)

a = 2ranion

2ranion + 2rcation = 2 2ranion

ranion + rcation = 2ranion rcation = ( 2 −1)ranion

2ranion + 2rcation = 2a

4140anion

cation .rr

=

٧٧Chapter 3 -

Rock Salt StructureSame concepts can be applied to ionic solids

in general. Example: NaCl (rock salt) structure

rNa = 0.102 nm

rNa/rCl = 0.564

∴ cations prefer OH sites


rCl = 0.181 nm

٧٨Chapter 3 -

MgO and FeOMgO and FeO also have the NaCl

structureO2- rO = 0.140 nm

Mg2+ rMg = 0.072 nm

rMg/rO = 0.514

∴ cations prefer OH sites

So each oxygen has 6 neighboring Mg2+


٧٩Chapter 3 -

• On the basis of ionic radii, what crystal structurewould you predict for FeO?

• Answer:

550014000770

anion

cation

...

rr

=

=

based on this ratio,--coord # = 6--structure = NaCl

Data from Table 12.3, Callister 7e.

Example: Predicting Structure of FeO

Ionic radius (nm)0.0530.0770.0690.100

0.1400.1810.133

Cation

Anion

Al3+

Fe2+

Fe3+

Ca2+

O2-

Cl-

F-

٨٠Chapter 3 -

AX Crystal StructuresAX–Type Crystal Structures include NaCl, CsCl, and

zinc blende

939.0181.0170.0

Cl

Cs ==−

+

rr


Cesium Chloride structure:

∴ cubic sites preferred

So each Cs+ has 8 neighboring Cl-

٨١Chapter 3 -

AX Crystal Structures

So each Zn2+ has 4 neighboring O2-

Zinc Blende structure


?? 529.0140.0074.0

2

2

O

ZnHO

rr

⇒==−

+

• Size arguments predict Zn2+

in OH sites, • In observed structure Zn2+

in TD sites

• Why is Zn2+ in TD sites?– bonding hybridization of

zinc favors TD sites

Ex: ZnO, ZnS, SiC

٨٢Chapter 3 -


Figure 3.33 (a) The zinc blende unit cell, (b) plan view.

٨٣Chapter 3 -

AX2 Crystal Structures

Fluoritestructure

• Calcium Fluorite (CaF2)• cations in cubic sites

• UO2, ThO2, ZrO2, CeO2

• antifluorite structure –cations and anions reversed


٨٤Chapter 3 -

ABX3 Crystal Structures

• Perovskite

Ex: complex oxide BaTiO3


٨٥Chapter 3 -

o Covalently bonded materials frequently have complex structures in order to satisfy the directional restraints imposed by the bonding.

o Diamond cubic (DC) - A special type of face-centered cubic crystal structure found in carbon, silicon, and other covalently bonded materials.

Covalent Structures

٨٦Chapter 3 -

(c) 2003 Brooks/C

ole Publishing / Thomson Learning

Figure . (a) Tetrahedron and (b) the diamond cubic (DC) unit cell. This open structure is produced because of the requirements of covalent bonding.

٨٧Chapter 3 -

(c) 2003 Brooks/C

ole Publishing / Thomson

Learning

Figure . The silicon-oxygen tetrahedron and the resultant β-cristobalite form of silica.

٨٨Chapter 3 -

X-Ray Diffraction

• Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation.

• Can’t resolve spacings < λ• Spacing is the distance between parallel planes of

atoms.

٨٩Chapter 3 -

X-Rays to Determine Crystal Structure

X-ray intensity (from detector)

θ

θc

d =nλ

2 sin θc

Measurement of critical angle, θc, allows computation of planar spacing, d.

• Incoming X-rays diffract from crystal planes.


reflections must be in phase for a detectable signal

spacing between planes

d

incoming

X-rays

outgo

ing X

-rays

detector

θλ

θextra distance travelledby wave “2”

“1”

“2”

“1”

“2”

٩٠Chapter 3 -

X-Ray Diffraction Pattern


(110)

(200)

(211)

z

x

ya b

c

Diffraction angle 2θ

Diffraction pattern for polycrystalline α-iron (BCC)

Inte

nsity

(rel

ativ

e)

z

x

ya b

cz

x

ya b

c

٩١Chapter 3 -

• 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).

SUMMARY

• Common metallic crystal structures are FCC, BCC, and HCP. Coordination number and atomic packing factorare the same for both FCC and HCP crystal structures.

• 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.

٩٢Chapter 3 -

• Some materials can have more than one crystal structure. This is referred to as polymorphism (or allotropy).

SUMMARY

• 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.

• X-ray diffraction is used for crystal structure and interplanar spacing determinations.

٩٣Chapter 3 -


• What types of defects arise in solids?

• Can the number and type of defects be variedand controlled?

• How do defects affect material properties?

• Are defects undesirable?

IMPERFECTIONS IN SOLIDS

• What are the solidification mechanisms?

٩٤Chapter 3 -

Imperfections in Solids

There is no such thing as a perfect crystal. • What are these imperfections? • Why are they important?

Many of the important properties of materials are due to the presence of imperfections.

٩٥Chapter 3 -

• Vacancy atoms• Interstitial atoms• Substitutional atoms

Point defects

Types of Imperfections

• Dislocations Line defects

• Grain Boundaries Area defects

٩٦Chapter 3 -

• Vacancies:-vacant atomic sites in a structure.

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

Point Defects

Vacancydistortion of planes

self-interstitial

distortion of planes

٩٧Chapter 3 -

Boltzmann's constant(1.38 x 10 -23 J/atom-K) (8.62 x 10-5 eV/atom-K)

NvN

= exp−QvkT

No. of defects

No. of potential defect sites.

Activation energy

Temperature

Each lattice site is a potential vacancy site

• Equilibrium concentration varies with temperature!

Equilibrium Concentration:Point Defects

٩٨Chapter 3 -

• Find the equil. # of vacancies in 1 m3 of Cu at 1000°C.• Given:

ACu = 63.5 g/molρ = 8.4 g/cm3

Qv = 0.9 eV/atom NA = 6.02 x 1023 atoms/mol

Estimating Vacancy Concentration

For 1 m3 , N =NAACu

ρ x x 1 m3= 8.0 x 1028 sites8.62 x 10-5 eV/atom-K

0.9 eV/atom

1273K

NvN

= exp−QvkT

= 2.7 x 10-4

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

٩٩Chapter 3 -

• Low energy electronmicroscope view ofa (110) surface of NiAl.

• Increasing T causessurface island ofatoms to grow.

• Why? The equil. vacancyconc. increases via atommotion from the crystalto the surface, where

they join the island.Reprinted with permission from Nature (K.F. McCarty, J.A. Nobel, and N.C. Bartelt, "Vacancies inSolids and the Stability of Surface Morphology",Nature, Vol. 412, pp. 622-625 (2001). Image is5.75 µm by 5.75 µm.) Copyright (2001) Macmillan Publishers, Ltd.

Observing Equilibrium Vacancy Conc.

Island grows/shrinks to maintain equil. vancancy conc. in the bulk.

١٠٠Chapter 3 -

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

• Solid solution of B in A plus particles of a newphase (usually for a larger amount of B)

OR

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

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

Second phase particle--different composition--often different structure.

Point Defects in Alloys

١٠١Chapter 3 -

Imperfections in Solids• Specification of composition

– weight percent 100x 21

11 mm

mC+

=

m1 = mass of component 1

100x 21

1'1

mm

m

nnnC+

=

nm1 = number of moles of component 1

– atom percent

١٠٢Chapter 3 -

• are line defects,• slip between crystal planes result when dislocations move,• produce permanent (plastic) deformation.

Dislocations:

Schematic of Zinc (HCP):• before deformation • after tensile elongation

slip steps

Line Defects


١٠٣Chapter 3 -


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 ⊥ to dislocation line

• Screw dislocation:– spiral planar ramp resulting from shear deformation– b || to dislocation line

Burger’s vector, b: measure of lattice distortion

١٠٤Chapter 3 -


Fig. 4.3, Callister 7e.

Edge Dislocation

١٠٥Chapter 3 -

• Dislocation motion requires the successive bumpingof a half plane of atoms (from left to right here).

• Bonds across the slipping planes are broken andremade in succession.

Atomic view of edgedislocation motion fromleft to right as a crystalis sheared.


Motion of Edge Dislocation

١٠٦Chapter 3 -


Screw Dislocation


Burgers vector b

Dislocationline

b

(a)(b)

Screw Dislocation

١٠٧Chapter 3 -

Edge, Screw, and Mixed Dislocations


Edge

Screw

Mixed

١٠٨Chapter 3 -


Dislocations are visible in electron micrographs


١٠٩Chapter 3 -

Grain boundaries...• are imperfections,• are more susceptible

to etching,• may be revealed as

dark lines,• change in crystal

orientation across boundary. Adapted from Fig. 4.14(a)

and (b), Callister 7e.(Fig. 4.14(b) is courtesyof L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC [now the National Institute of Standards and Technology, Gaithersburg, MD].)

Optical Microscopy

Fe-Cr alloy(b)

grain boundarysurface groove

polished surface

(a)

١١٠Chapter 3 -

• Point, Line, and Area defects exist in solids.

• The number and type of defects can be variedand controlled (e.g., T controls vacancy conc.)

• Defects affect material properties (e.g., grainboundaries control crystal slip).

• Defects may be desirable or undesirable(e.g., dislocations may be good or bad, dependingon whether plastic deformation is desirable or not.)

Summary

١١١Chapter 3 -١١١

ISSUES TO ADDRESS...• Stress and strain: What are they and why are

they used instead of load and deformation?• Elastic behavior: When loads are small, how much

deformation occurs? What materials deform least?• Plastic behavior: At what point does permanent

deformation occur? What materials are most resistant to permanent deformation?

• Toughness and ductility: What are they and howdo we measure them?

Mechanical Properties

١١٢Chapter 3 -١١٢

Terminology for Mechanical Propertiesq Stress - Force or load per unit area of cross-section over

which the force or load is acting.q Strain - Elongation change in dimension per unit length.q Young’s modulus - The slope of the linear part of the stress-strain curve in the elastic region, same as modulus of elasticity.q Shear modulus (G) - The slope of the linear part of the shear stress-shear strain curve.q Viscosity ( ) - Measure of resistance to flow, defined as the ratio of shear stress to shear strain rate (units Poise or Pa-s).q Thixotropic behavior - Materials that show shear thinning and also an apparent viscosity that at a constant rate of shear decreases with time.

η

١١٣Chapter 3 -١١٣

(photo courtesy P.M. Anderson)Canyon Bridge, Los Alamos, NM

oσ = F

A

• Simple compression:

Note: compressivestructure member(σ < 0 here).(photo courtesy P.M. Anderson)

OTHER COMMON STRESS STATES (1)

Ao

Balanced Rock, Arches National Park

١١٤Chapter 3 -١١٤

∴ Stress has units:N/m2 or lbf/in2

Engineering Stress• Shear stress, τ:

Area, A

Ft

Ft

Fs

F

F

Fs

τ = FsAo

• Tensile stress, σ:

original area before loading

Area, A

Ft

Ft

σ = FtAo

2f

2mNor

inlb=

١١٥Chapter 3 -١١٥

• Tensile strain: • Lateral strain:

• Shear strain:

Strain is alwaysdimensionless.

Engineering Strain

θ

90º

90º - θy

∆x θγ = ∆x/y = tan

ε = δLo

−δεL = L

wo

Adapted from Fig. 6.1 (a) and (c), Callister 7e.

δ/2

δL/2

Lowo

١١٦Chapter 3 -١١٦

١١٧Chapter 3 -١١٧

Figure (a) Tensile, compressive, shear and bending stresses. (b) Illustration showing how Young’s modulus is defined for elastic material. (c) For nonlinear materials, we use the slope of a tangent as a variable quantity that replaces the Young’s modulus constant

(c)2

003

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١١٨Chapter 3 -١١٨

Linear Elastic Properties• Modulus of Elasticity, E:

(also known as Young's modulus)

• Hooke's Law:σ = E ε σ

Linear-elastic

E

ε

F

Fsimple tension test

١١٩Chapter 3 -١١٩

• Elastic Shearmodulus, G:

τG

γτ = G γ

Other Elastic Properties

simpletorsiontest

M

M

• Special relations for isotropic materials:

2(1 + ν)EG =

3(1 − 2ν)EK =

١٢٠Chapter 3 -١٢٠

Poisson's ratio, ν• Poisson's ratio, ν:

Units:E: [GPa] or [psi]ν: dimensionless

–ν > 0.50 density increases

–ν < 0.50 density decreases (voids form)

εL

ε

-ν

εν = − L

ε

metals: ν ~ 0.33ceramics: ν ~ 0.25polymers: ν ~ 0.40

١٢١Chapter 3 -١٢١

Stress-Strain Testing• Typical tensile test

machine

Adapted from Fig. 6.3, Callister 7e. (Fig. 6.3 is taken from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.)

specimenextensometer

• Typical tensile specimen


gauge length

١٢٢Chapter 3 -١٢٢

Tensile Strength, TS

• Metals: occurs when noticeable necking starts.• Polymers: occurs when polymer backbone chains are

aligned and about to break.


σy

strain

Typical response of a metal

F = fracture or ultimate strength

Neck – acts as stress concentrator

engi

neer

ing

TSst

ress

engineering strain

١٢٣Chapter 3 -١٢٣

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Figure 6.12 Localized deformation of a ductile material during a tensile test produces a necked region. The micrograph shows necked region in a fractured sample

١٢٤Chapter 3 -١٢٤

١٢٥Chapter 3 -١٢٥


Figure :The stress-strain curve for an aluminum alloy

١٢٦Chapter 3 -١٢٦

Example :Tensile Testing of Aluminum Alloy

Convert the change in length data in Table 6-1 to engineering stress and strain and plot a stress-strain curve.

١٢٧Chapter 3 -١٢٧

(c)2

003

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Figure: Tensile stress-strain curves for different materials. Note that these are qualitative

١٢٨Chapter 3 -١٢٨


Figure: (a) Determining the 0.2% offset yield strength in gray cast ion, and (b) upper and lower yield point behavior in a low-carbon steel

١٢٩Chapter 3 -١٢٩

• Stress at which noticeable plastic deformation hasoccurred.

when εp = 0.002

Yield Strength, σy

σy = yield strength

Note: for 2 inch sample

ε = 0.002 = ∆z/z

∴ ∆z = 0.004 in

Adapted from Fig. 6.10 (a),Callister 7e.

tensile stress, σ

engineering strain, ε

σy

εp = 0.002

١٣٠Chapter 3 -١٣٠

Figure 6.13 Typical yield strength values for different engineered materials. (Source: Reprinted fromEngineering Materials I, Second Edition, M.F. Ashby and D.R.H. Jones, 1996, Fig. 8-12, p. 85. Copyright ©Butterworth-Heinemann. Reprinted with permission from Elsevier Science.)

١٣١Chapter 3 -١٣١

١٣٢Chapter 3 -١٣٢

Figure : Range of elastic moduli for different engineered materials. (Source: Reprinted from Engineering Materials I, Second Edition, M.F. Ashby and D.R.H. Jones, 1996, Fig. 3-5, p. 35, Copyright © 1996 Butterworth-Heinemann. Reprinted with permission from Elsevier Science.)

١٣٣Chapter 3 -١٣٣

True Stress & StrainNote: S.A. changes when sample

stretched

• True stress

• True Strain

iT AF=σ

( )oiT llln=ε( )( )ε+=ε

ε+σ=σ1ln1

T

T


١٣٤Chapter 3 -١٣٤

Example : Young’s Modulus of Aluminum Alloy

From the data in Example 6.1, calculate the modulus of elasticity of the aluminum alloy. Use the modulus to determine the length after deformation of a bar of initial length of 50 in. Assume that a level of stress of 30,000 psi is applied.

Example SOLUTION

١٣٥Chapter 3 -١٣٥

Example: True Stress and True Strain Calculation

Compare engineering stress and strain with true stress and strain for the aluminum alloy in Example 6.1 at (a) the maximum load and (b) fracture. The diameter at maximum load is 0.497 in. and at fracture is 0.398 in.

Example 6.5 SOLUTION

١٣٦Chapter 3 -١٣٦

Example : SOLUTION (Continued)

١٣٧Chapter 3 -١٣٧

Hardening

• Curve fit to the stress-strain response:

σT = K εT( )n“true” stress (F/A) “true” strain: ln(L/Lo)

hardening exponent:n = 0.15 (some steels) to n = 0.5 (some coppers)

• An increase in σy due to plastic deformation.σ

ε

large hardening

small hardeningσy0

σy1

١٣٨Chapter 3 -١٣٨

Resilience, Ur

• Ability of a material to store energy – Energy stored best in elastic region

If we assume a linear stress-strain curve this simplifies to


yyr 21U εσ≅

∫ε

εσ= y dU r 0

١٣٩Chapter 3 -١٣٩

• Energy to break a unit volume of material• Approximate by the area under the stress-strain

curve.

Toughness

Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy

very small toughness (unreinforced polymers)

Engineering tensile strain, ε

Engineeringtensile stress, σ

small toughness (ceramics)

large toughness (metals)


١٤٠Chapter 3 -١٤٠

Impact Testing

final height initial height

• Impact loading:-- severe testing case-- makes material more brittle-- decreases toughness

Adapted from Fig. 8.12(b), Callister 7e. (Fig. 8.12(b) is adapted from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, John Wiley and Sons, Inc. (1965) p. 13.)

(Charpy)

١٤١Chapter 3 -١٤١

• Plastic tensile strain at failure:


Ductility

• Another ductility measure: 100xA

AARA%o

fo -=

x 100L

LLEL%o

of −=

Engineering tensile strain, ε

Engineeringtensile stress, σ

smaller %EL

larger %ELLf

Ao AfLo

١٤٢Chapter 3 -١٤٢


Figure : The stress-strain behavior of brittle materials compared with that of more ductile materials

١٤٣Chapter 3 -١٤٣


Figure : The effect of temperance (a) on the stress-strain curve and (b) on the tensile properties of an aluminum alloy

١٤٤Chapter 3 -١٤٤

Example : Ductility of an Aluminum Alloy

The aluminum alloy in Example 6.1 has a final length after failure of 2.195 in. and a final diameter of 0.398 in. at the fractured surface. Calculate the ductility of this alloy.

Example SOLUTION

١٤٥Chapter 3 -١٤٥

Hardness• Resistance to permanently indenting the surface.• Large hardness means:

--resistance to plastic deformation or cracking incompression.

--better wear properties.

e.g., 10 mm sphere

apply known force measure size of indent after removing load

dDSmaller indents mean larger hardness.

increasing hardness

most plastics

brasses Al alloys

easy to machine steels file hard

cutting tools

nitridedsteels diamond

١٤٦Chapter 3 -١٤٦

Hardness: Measurement

• Rockwell– No major sample damage– Each scale runs to 130 but only useful in range

20-100. – Minor load 10 kg– Major load 60 (A), 100 (B) & 150 (C) kg

• A = diamond, B = 1/16 in. ball, C = diamond

• HB = Brinell Hardness– TS (psia) = 500 x HB– TS (MPa) = 3.45 x HB

١٤٧Chapter 3 -١٤٧

Hardness: MeasurementTable

١٤٨Chapter 3 -١٤٨

ISSUES TO ADDRESS...• How do flaws in a material initiate failure?• How is fracture resistance quantified; how do different

material classes compare?• How do we estimate the stress to fracture?• How do loading rate, loading history, and temperature

affect the failure stress?

Ship-cyclic loadingfrom waves.

Computer chip-cyclicthermal loading.

Hip implant-cyclicloading from walking.

Adapted from Fig. 22.30(b), Callister 7e.(Fig. 22.30(b) is courtesy of National Semiconductor Corporation.)


Mechanical Failure

Adapted from chapter-opening photograph, Chapter 8, Callister 7e. (by Neil Boenzi, The New York Times.)

١٤٩Chapter 3 -١٤٩

Fracture mechanisms

• Ductile fracture– Occurs with plastic deformation

• Brittle fracture– Little or no plastic deformation– Catastrophic

١٥٠Chapter 3 -١٥٠

Ductile vs Brittle FailureVery

DuctileModerately

Ductile BrittleFracturebehavior:

Large Moderate%AR or %EL Small• Ductilefracture is usuallydesirable!


• Classification:

Ductile:warning before

fracture

Brittle:No

warning

١٥١Chapter 3 -١٥١

• Ductile failure:--one piece--large deformation

Figures from V.J. Colangelo and F.A. Heiser, Analysis of Metallurgical Failures(2nd ed.), Fig. 4.1(a) and (b), p. 66 John Wiley and Sons, Inc., 1987. Used with permission.

Example: Failure of a Pipe

• Brittle failure:--many pieces--small deformation

١٥٢Chapter 3 -١٥٢

• Evolution to failure:

• Resultingfracturesurfaces(steel)

50 mm

particlesserve as voidnucleationsites.

50 mm

From V.J. Colangelo and F.A. Heiser, Analysis of Metallurgical Failures (2nd ed.), Fig. 11.28, p. 294, John Wiley and Sons, Inc., 1987. (Orig. source: P. Thornton, J. Mater. Sci., Vol. 6, 1971, pp. 347-56.)

100 mmFracture surface of tire cord wire loaded in tension. Courtesy of F. Roehrig, CC Technologies, Dublin, OH. Used with permission.

Moderately Ductile Failurenecking

σ

void nucleation

void growth and linkage

shearing at surface fracture

١٥٣Chapter 3 -١٥٣

Ductile vs. Brittle Failure


cup-and-cone fracture brittle fracture

١٥٤Chapter 3 -١٥٤

Fracture Mechanics

o Fracture mechanics - The study of a material’s ability to withstand stress in the presence of a flaw.

o Fracture toughness - The resistance of a material to failure in the presence of a flaw.

١٥٥Chapter 3 -١٥٥

(c)2

003

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Figure : The fracture toughness Kc of a 3000,000psi yield strength steel decreases with increasing thickness, eventually leveling off at the plane strain fracture toughness Klc

١٥٦Chapter 3 -١٥٦

Design of a Nondestructive Test

A large steel plate used in a nuclear reactor has a plane strain fracture toughness of 80,000 psi and is exposed to a stress of 45,000 psi during service. Design a testing or inspection procedure capable of detecting a crack at the edge of the plate before the crack is likely to grow at a catastrophic rate.

SOLUTION

.in

١٥٧Chapter 3 -١٥٧

Design of a Ceramic Support

Design a supporting 3-in.-wide plate made of sialon, which has a fracture toughness of 9,000 psi , that will withstand a tensile load of 40,000 lb. The part is to be nondestructively tested to assure that no flaws are present that might cause failure.

.in

١٥٨Chapter 3 -١٥٨

Flaws are Stress Concentrators!Results from crack

propagation• Griffith Crack

where ρt = radius of curvatureσo = applied stressσm = stress at crack tip

ot

/

tom Ka

σ=

ρ

σ=σ21

2

ρt


١٥٩Chapter 3 -١٥٩

(c)2

003

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Figure:Schematic diagram of the Griffith flaw in a ceramic

١٦٠Chapter 3 -١٦٠

Concentration of Stress at Crack Tip

١٦١Chapter 3 -١٦١

Crack PropagationCracks propagate due to sharpness of crack

tip • A plastic material deforms at the tip,

“blunting” the crack.deformed region

brittle

Energy balance on the crack• Elastic strain energy-

• energy stored in material as it is elastically deformed• this energy is released when the crack propagates• creation of new surfaces requires energy

plastic

١٦٢Chapter 3 -١٦٢

When Does a Crack Propagate?Crack propagates if above critical

stress

• where– E = modulus of elasticity– γs = specific surface energy– a = one half length of internal crack– Kc = σc/σ0

For ductile => replace γs by γs + γpwhere γp is plastic deformation energy

212 /s

c aE

πγ

=σi.e., σm > σc

or Kt > Kc

١٦٣Chapter 3 -١٦٣

Fatigue• Fatigue = failure under cyclic stress.

• Stress varies with time.-- key parameters are S, σm, and

frequency

σmax

σmin

σ

time

σmS

• Key points: Fatigue...--can cause part failure, even though σmax < σc.--causes ~ 90% of mechanical engineering failures.

Adapted from Fig. 8.18, Callister 7e. (Fig. 8.18 is from Materials Science in Engineering, 4/E by Carl. A. Keyser, Pearson Education, Inc., Upper Saddle River, NJ.)tension on bottom

compression on top

countermotor

flex coupling

specimen

bearing bearing

١٦٤Chapter 3 -١٦٤

• Fatigue limit, Sfat:--no fatigue if S < Sfat


Fatigue Design Parameters

Sfat

case for steel (typ.)

N = Cycles to failure103 105 107 109

unsafe

safe

S = stress amplitude

• Sometimes, thefatigue limit is zero!


case for Al (typ.)

N = Cycles to failure103 105 107 109

unsafe

safe

S = stress amplitude

١٦٥Chapter 3 -١٦٥

CreepSample deformation at a constant stress (σ) vs.

time

Adapted fromFig. 8.28, Callister 7e.

Primary Creep: slope (creep rate) decreases with time.

Secondary Creep: steady-statei.e., constant slope.

Tertiary Creep: slope (creep rate) increases with time, i.e. acceleration of rate.

σσ,ε

0 t

١٦٦Chapter 3 -١٦٦

• Occurs at elevated temperature, T > 0.4 Tm

Adapted from Figs. 8.29, Callister 7e.

Creep

elastic

primarysecondary

tertiary

١٦٧Chapter 3 -١٦٧

Strategies for Strengthening: 1: Reduce Grain Size

• Grain boundaries arebarriers to slip.

• Barrier "strength"increases withIncreasing angle of misorientation.

• Smaller grain size:more barriers to slip.

• Hall-Petch Equation: 21 /yoyield dk −+σ=σ

Adapted from Fig. 7.14, Callister 7e.(Fig. 7.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

١٦٨Chapter 3 -١٦٨

• Impurity atoms distort the lattice & generate stress.• Stress can produce a barrier to dislocation motion.

Strategies for Strengthening: 2: Solid Solutions

• Smaller substitutionalimpurity

Impurity generates local stress at Aand B that opposes dislocation motion to the right.

A

B

• Larger substitutionalimpurity

Impurity generates local stress at Cand D that opposes dislocation motion to the right.

C

D

١٦٩Chapter 3 -١٦٩

Stress Concentration at Dislocations


١٧٠Chapter 3 -١٧٠

Strengthening by Alloying• small impurities tend to concentrate at

dislocations• reduce mobility of dislocation ∴ increase

strength


١٧١Chapter 3 -١٧١

Strengthening by alloying

• large impurities concentrate at dislocations on low density side


١٧٢Chapter 3 -١٧٢

Ex: Solid SolutionStrengthening in Copper

• Tensile strength & yield strength increase with wt% Ni.

• Empirical relation:

• Alloying increases σy and TS.

21 /y C~σ

Adapted from Fig. 7.16 (a) and (b), Callister 7e.

Tens

ile s

treng

th (M

Pa)

wt.% Ni, (Concentration C)

200

300

400

0 10 20 30 40 50 Yiel

d st

reng

th (M

Pa)

wt.%Ni, (Concentration C)

60

120

180

0 10 20 30 40 50

١٧٣Chapter 3 -١٧٣

Strategies for Strengthening: 3: Cold Work (%CW)

• Room temperature deformation.• Common forming operations change the cross

sectional area:


-Forging

Ao Ad

force

dieblank

force-Drawing

tensile force

AoAddie

die

-Extrusion

ram billet

container

containerforce die holder

die

Ao

Adextrusion

100 x %o

doA

AACW −=

-Rolling

roll

AoAd

roll

١٧٤Chapter 3 -١٧٤

• Ti alloy after cold working:

• Dislocations entanglewith one anotherduring cold work.

• Dislocation motionbecomes more difficult.

Adapted from Fig. 4.6, Callister 7e.(Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.)

Dislocations During Cold Work

0.9 µm

١٧٥Chapter 3 -١٧٥

• What is the tensile strength &ductility after cold working?

Adapted from Fig. 7.19, Callister 7e. (Fig. 7.19 is adapted from Metals Handbook: Properties and Selection: Iron and Steels, Vol. 1, 9th ed., B. Bardes (Ed.), American Society for Metals, 1978, p. 226; and Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th ed., H. Baker (Managing Ed.), American Society for Metals, 1979, p. 276 and 327.)

%6.35100 x % 2

22=

π

π−π=

o

dor

rrCW

Cold Work Analysis

% Cold Work

100

300

500

700

Cu

200 40 60

yield strength (MPa)

σy = 300MPa

300MPa

% Cold Work

tensile strength (MPa)

200Cu

0

400

600

800

20 40 60

ductility (%EL)

% Cold Work

20

40

60

20 40 6000

Cu

Do =15.2mm

Cold Work

Dd =12.2mm

Copper

340MPa

TS = 340MPa

7%

%EL = 7%

١٧٦Chapter 3 -١٧٦


• How does diffusion occur?

• Why is it an important part of processing?

• How can the rate of diffusion be predicted forsome simple cases?

• How does diffusion depend on structureand temperature?

Diffusion in Solids

١٧٧Chapter 3 -١٧٧

Diffusion

Diffusion - Mass transport by atomic motion

Mechanisms• Gases & Liquids – random (Brownian)

motion• Solids – vacancy diffusion or interstitial

diffusion

١٧٨Chapter 3 -١٧٨

• Interdiffusion: In an alloy, atoms tend to migratefrom regions of high conc. to regions of low conc.

Initially

Adapted from Figs. 5.1 and 5.2, Callister7e.

Diffusion

After some time

١٧٩Chapter 3 -١٧٩

• Self-diffusion: In an elemental solid, atomsalso migrate.

Label some atoms After some time

Diffusion

A

B

C

DA

B

C

D

١٨٠Chapter 3 -١٨٠

Diffusion MechanismsVacancy Diffusion:• atoms exchange with vacancies• applies to substitutional impurities atoms • rate depends on:

--number of vacancies--activation energy to exchange.

increasing elapsed time

١٨١Chapter 3 -١٨١

• Simulation of interdiffusionacross an interface:

• Rate of substitutionaldiffusion depends on:--vacancy concentration--frequency of jumping.


Diffusion Simulation

١٨٢Chapter 3 -١٨٢

Diffusion Mechanisms• Interstitial diffusion – smaller

atoms can diffuse between atoms.

More rapid than vacancy diffusionAdapted from Fig. 5.3 (b), Callister 7e.

١٨٣Chapter 3 -١٨٣

Adapted from chapter-opening photograph, Chapter 5, Callister 7e. (Courtesy ofSurface Division, Midland-Ross.)

• Case Hardening:--Diffuse carbon atoms

into the host iron atomsat the surface.

--Example of interstitialdiffusion is a casehardened gear.

• Result: The presence of C atoms makes iron (steel) harder.

Processing Using Diffusion

١٨٤Chapter 3 -١٨٤

• Doping silicon with phosphorus for n-type semiconductors:• Process:

3. Result: Dopedsemiconductorregions.

silicon

Processing Using Diffusion

magnified image of a computer chip

0.5mm

light regions: Si atoms

light regions: Al atoms

2. Heat it.

1. Deposit P richlayers on surface.

silicon


١٨٥Chapter 3 -١٨٥

Diffusion• How do we quantify the amount or rate of

diffusion?

• Measured empirically– Make thin film (membrane) of known surface area– Impose concentration gradient– Measure how fast atoms or molecules diffuse through

the membrane

( )( ) smkgor

scmmol

timearea surfacediffusing mass) (or molesFlux 22=≡≡J

dtdM

Al

AtMJ ==

M =mass

diffusedtime

J ∝ slope

١٨٦Chapter 3 -١٨٦

Steady-State Diffusion

dxdCDJ −=

Fick’s first law of diffusionC1

C2

x

C1

C2

x1 x2

D ≡ diffusion coefficient

Rate of diffusion independent of timeFlux proportional to concentration gradient =

dxdC

12

12 linear ifxxCC

xC

dxdC

−−

=∆∆

≅

١٨٧Chapter 3 -١٨٧

Example: Chemical Protective Clothing (CPC)

• Methylene chloride is a common ingredient of paint removers. Besides being an irritant, it also may be absorbed through skin. When using this paint remover, protective gloves should be worn.

• If butyl rubber gloves (0.04 cm thick) are used, what is the diffusive flux of methylene chloride through the glove?

• Data:– diffusion coefficient in butyl rubber:

D = 110 x10-8 cm2/s– surface concentrations:

C2 = 0.02 g/cm3C1 = 0.44 g/cm3

١٨٨Chapter 3 -١٨٨

scmg 10 x 16.1

cm) 04.0()g/cm 44.0g/cm 02.0(/s)cm 10 x 110( 2

5-33

28- =−

−=J

Example (cont).

12

12- xxCCD

dxdCDJ

−−

−≅=

Dtb 6

2l=

gloveC1

C2

skinpaintremover

x1 x2

• Solution – assuming linear conc. gradient

D = 110x10-8 cm2/s

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

x2 – x1 = 0.04 cm

Data:

١٨٩Chapter 3 -١٨٩

Diffusion and Temperature

• Diffusion coefficient increases with increasing T.

D = Do exp

−

Qd

RT

= pre-exponential [m2/s]= diffusion coefficient [m2/s]

= activation energy [J/mol or eV/atom] = gas constant [8.314 J/mol-K]= absolute temperature [K]

DDo

Qd

RT

١٩٠Chapter 3 -١٩٠

Diffusion and Temperature

Adapted from Fig. 5.7, Callister 7e. (Date for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)

D has exponential dependence on T

Dinterstitial >> DsubstitutionalC in α-FeC in γ-Fe

Al in AlFe in α-FeFe in γ-Fe

1000K/T

D (m2/s) C in α-Fe

C in γ-Fe

Al in Al

Fe in α-Fe

Fe in γ-Fe

0.5 1.0 1.510-20

10-14

10-8T(°C)15

00

1000

600

300

١٩١Chapter 3 -١٩١

Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are

D(300ºC) = 7.8 x 10-11 m2/sQd = 41.5 kJ/mol

What is the diffusion coefficient at 350ºC?

−=

−=

101

202

1lnln and 1lnlnTR

QDDTR

QDD dd

−−==−∴

121

212

11lnlnln TTR

QDDDD d

transform data

D

Temp = T

ln D

1/T

١٩٢Chapter 3 -١٩٢

Example (cont.)

−

−= −

K 5731

K 6231

K-J/mol 314.8J/mol 500,41exp /s)m 10 x 8.7( 211

2D

−−=

1212

11exp TTR

QDD d

T1 = 273 + 300 = 573KT2 = 273 + 350 = 623K

D2 = 15.7 x 10-11 m2/s

١٩٣Chapter 3 -١٩٣

Non-steady State Diffusion

• The concentration of diffucing species is a function of both time and position C = C(x,t)

• In this case Fick’s Second Law is used

2

2

xCD

tC

∂∂

=∂∂Fick’s Second Law

١٩٤Chapter 3 -١٩٤

Non-steady State Diffusion


B.C. at t = 0, C = Co for 0 ≤ x ≤ ∞

at t > 0, C = CS for x = 0 (const. surf. conc.)

C = Co for x = ∞

• Copper diffuses into a bar of aluminum.

pre-existing conc., Co of copper atoms

Surface conc., C of Cu atoms bars

Cs

١٩٥Chapter 3 -١٩٥

Diffusion FASTER for...

• open crystal structures

• materials w/secondarybonding

• smaller diffusing atoms

• lower density materials

Diffusion SLOWER for...

• close-packed structures

• materials w/covalentbonding

• larger diffusing atoms

• higher density materials

Summary

١٩٦Chapter 3 -

ISSUES TO ADDRESS...• When we combine two elements...

what equilibrium state do we get?• In particular, if we specify...

--a composition (e.g., wt% Cu - wt% Ni), and--a temperature (T)

then...How many phases do we get?What is the composition of each phase?How much of each phase do we get?

Chapter : Phase Diagrams

Phase BPhase A

Nickel atomCopper atom

١٩٧Chapter 3 -

Phase Equilibria: Solubility LimitIntroduction

– Solutions – solid solutions, single phase– Mixtures – more than one phase

• Solubility Limit:Max concentration forwhich only a single phase solution occurs.

Question: What is thesolubility limit at 20°C?

Answer: 65 wt% sugar.If Co < 65 wt% sugar: syrupIf Co > 65 wt% sugar: syrup + sugar. 65

Sucrose/Water Phase Diagram

Pure

Su

gar

Tem

pera

ture

(°C

)

0 20 40 60 80 100Co =Composition (wt% sugar)

L(liquid solution

i.e., syrup)

Solubility Limit L

(liquid) + S

(solid sugar)20

40

60

80

100

Pure

W

ater


١٩٨Chapter 3 -

• Components:The elements or compounds which are present in the mixture

(e.g., Al and Cu)• Phases:

The physically and chemically distinct material regionsthat result (e.g., α and β).

Aluminum-CopperAlloy

Components and Phases

α (darker phase)

β (lighter phase)


١٩٩Chapter 3 -

Effect of T & Composition (Co)• Changing T can change # of phases:


D (100°C,90)2 phases

B (100°C,70)1 phase

path A to B.• Changing Co can change # of phases: path B to D.

A (20°C,70)2 phases

70 80 1006040200

Tem

pera

ture

(°C

)

Co =Composition (wt% sugar)

L(liquid solution

i.e., syrup)

20

100

40

60

80

0

L(liquid)

+ S

(solid sugar)

water-sugarsystem

٢٠٠Chapter 3 -

Phase Equilibria

0.1278

1.8FCCCu0.124

61.9FCCNi

r (nm)electroneg

CrystalStructur

e

• Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume –Rothery rules) suggesting high mutual solubility.

Simple solution system (e.g., Ni-Cu solution)

• Ni and Cu are totally miscible in all proportions.

٢٠١Chapter 3 -

Phase Diagrams• Indicate phases as function of T, Co, and P. • For this course:

-binary systems: just 2 components.-independent variables: T and Co (P = 1 atm is almost always used).

• PhaseDiagramfor Cu-Nisystem

Adapted from Fig. 9.3(a), Callister 7e.(Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

• 2 phases:L (liquid)α (FCC solid solution)

• 3 phase fields: LL + αα

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

α(FCC solid solution)

L + αliquidus

solidus

٢٠٢Chapter 3 -

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

α(FCC solid solution)

L + αliquidus

solidusCu-Niphase

diagram

Phase Diagrams:# and types of phases

• Rule 1: If we know T and Co, then we know:--the # and types of phases present.

• Examples:A(1100°C, 60):

1 phase: α

B(1250°C, 35): 2 phases: L + α

Adapted from Fig. 9.3(a), Callister 7e.(Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991).

B(1

250°

C,3

5)A(1100°C,60)

٢٠٣Chapter 3 -

wt% Ni20

1200

1300

T(°C)

L (liquid)

α(solid)L + α

liquidus

solidus

30 40 50

L + α

Cu-Ni system

Phase Diagrams:composition of phases

• Rule 2: If we know T and Co, then we know:--the composition of each phase.

• Examples:TA A

35Co

32CL

At TA = 1320°C: Only Liquid (L) CL = Co ( = 35 wt% Ni)

At TB = 1250°C: Both α and LCL = C liquidus ( = 32 wt% Ni here) Cα = Csolidus ( = 43 wt% Ni here)

At TD = 1190°C: Only Solid ( α) Cα = Co ( = 35 wt% Ni)

Co = 35 wt% Ni

Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagramsof Binary Nickel Alloys, P. Nash (Ed.), ASM

International, Materials Park, OH, 1991.)

BTB

DTD

tie line

4Cα3

٢٠٤Chapter 3 -

• Rule 3: If we know T and Co, then we know:--the amount of each phase (given in wt%).

• Examples:

At TA: Only Liquid (L) WL = 100 wt%, Wα = 0

At TD: Only Solid ( α) WL = 0, Wα = 100 wt%

Co = 35 wt% Ni

Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.)

Phase Diagrams:weight fractions of phases

wt% Ni20

1200

1300

T(°C)

L (liquid)

α(solid)L + α

liquidus

solidus

30 40 50

L + α

Cu-Ni system

TA A

35Co

32CL

BTB

DTD

tie line

4Cα3

R S

At TB: Both α and L

% 7332433543 wt=

−−

=

= 27 wt%

WL = SR +S

Wα = RR +S

٢٠٥Chapter 3 -

• Tie line – connects the phases in equilibrium with each other - essentially an isotherm

The Lever Rule

How much of each phase?Think of it as a lever (teeter-totter)

ML Mα

R S

RMSM L ⋅=⋅α

L

L

LL

LL CC

CCSR

RWCCCC

SRS

MMMW

−−

=+

=−−

=+

=+

=α

αα

α

α

00

wt% Ni20

1200

1300

T(°C)

L (liquid)

α(solid)L + α

liquidus

solidus

30 40 50

L + αB

TB

tie line

CoCL Cα

SR


٢٠٦Chapter 3 -

wt% Ni20

1200

1300

30 40 501100

L (liquid)

α(solid)

L + α

L + α

T(°C)

A

35Co

L: 35wt%NiCu-Ni

system

• Phase diagram:Cu-Ni system.

• System is:--binary

i.e., 2 components:Cu and Ni.

--isomorphousi.e., completesolubility of onecomponent inanother; α phasefield extends from0 to 100 wt% Ni.


• ConsiderCo = 35 wt%Ni.

Ex: Cooling in a Cu-Ni Binary

46354332

α: 43 wt% Ni

L: 32 wt% Ni

L: 24 wt% Ni

α: 36 wt% Ni

Bα: 46 wt% NiL: 35 wt% Ni

C

D

E

24 36

٢٠٧Chapter 3 -

• Cα changes as we solidify.• Cu-Ni case:

• Fast rate of cooling:Cored structure

• Slow rate of cooling:Equilibrium structure

First α to solidify has Cα = 46 wt% Ni.Last α to solidify has Cα = 35 wt% Ni.

Cored vs Equilibrium Phases

First α to solidify: 46 wt% Ni

Uniform Cα:

35 wt% Ni

Last α to solidify: < 35 wt% Ni

٢٠٨Chapter 3 -

: Min. melting TE

2 componentshas a special compositionwith a min. melting T.


Binary-Eutectic Systems

• Eutectic transitionL(CE) α(CαE) + β(CβE)

• 3 single phase regions (L, α, β)

• Limited solubility: α: mostly Cu β: mostly Ag

• TE : No liquid below TE

• CE

composition

Ex.: Cu-Ag systemCu-Agsystem

L (liquid)

α L + α L+ββ

α + β

Co , wt% Ag20 40 60 80 1000

200

1200T(°C)

400

600

800

1000

CE

TE 8.0 71.9 91.2779°C

٢٠٩Chapter 3 -

L+αL+β

α + β

200

T(°C)

18.3

C, wt% Sn20 60 80 1000

300

100

L (liquid)

α 183°C61.9 97.8

β

• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...--the phases present: Pb-Sn

system

EX: Pb-Sn Eutectic System (1)

α + β--compositions of phases:

CO = 40 wt% Sn

--the relative amountof each phase:

150

40Co

11Cα

99Cβ

SR

Cα = 11 wt% SnCβ = 99 wt% Sn

Wα=Cβ - COCβ - Cα

= 99 - 4099 - 11 = 59

88 = 67 wt%

SR+S =

Wβ =CO - Cα

Cβ - Cα=R

R+S

= 2988

= 33 wt%= 40 - 1199 - 11


٢١٠Chapter 3 -

L+β

α + β

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

α βL+α

183°C

• For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find...--the phases present: Pb-Sn

system


EX: Pb-Sn Eutectic System (2)

α + L--compositions of phases:

CO = 40 wt% Sn

--the relative amountof each phase:

Wα =CL - CO

CL - Cα=

46 - 4046 - 17

=629 = 21 wt%

WL =CO - Cα

CL - Cα=

2329 = 79 wt%

40Co

46CL

17Cα

220SR

Cα = 17 wt% SnCL = 46 wt% Sn

٢١١Chapter 3 -

• Co < 2 wt% Sn• Result:

--at extreme ends--polycrystal of α grains

i.e., only one solid phase.


Microstructures in Eutectic Systems: I

0

L+ α200

T(°C)

Co, wt% Sn10

2

20Co

300

100

L

α

30

α+β

400

(room T solubility limit)

TE(Pb-SnSystem)

αL

L: Co wt% Sn

α: Co wt% Sn

٢١٢Chapter 3 -

• 2 wt% Sn < Co < 18.3 wt% Sn• Result:§ Initially liquid + α§ then α alone§ finally two phasesØ α polycrystalØ fine β-phase inclusions


Microstructures in Eutectic Systems: II

Pb-Snsystem

L + α

200

T(°C)

Co , wt% Sn10

18.3

200Co

300

100

L

α

30

α+ β

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

Lα

L: Co wt% Sn

αβ

α: Co wt% Sn

٢١٣Chapter 3 -

• Co = CE• Result: Eutectic microstructure (lamellar structure)

--alternating layers (lamellae) of α and β crystals.


Microstructures in Eutectic Systems: III

Adapted from Fig. 9.14, Callister 7e.160µm

Micrograph of Pb-Sneutectic microstructure

Pb-Snsystem

L + β

α + β

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L

α βL+α

183°C

40

TE

18.3

α: 18.3 wt%Sn

97.8

β: 97.8 wt% Sn

CE61.9

L: Co wt% Sn

٢١٤Chapter 3 -

Lamellar Eutectic Structure

Adapted from Figs. 9.14 & 9.15, Callister 7e.

٢١٥Chapter 3 -

• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: α crystals and a eutectic microstructure

Microstructures in Eutectic Systems: IV

18.3 61.9

SR

97.8

SR

primary αeutectic α

eutectic β

WL = (1- Wα) = 50 wt%

Cα = 18.3 wt% SnCL = 61.9 wt% Sn

SR + S

Wα= = 50 wt%

• Just above TE :

• Just below TE :Cα = 18.3 wt% SnCβ = 97.8 wt% Sn

SR + S

Wα= = 73 wt%

Wβ = 27 wt%Adapted from Fig. 9.16, Callister 7e.

Pb-Snsystem

L+β200

T(°C)

Co, wt% Sn

20 60 80 1000

300

100

L

α βL+α

40

α+β

TE

L: Co wt% Sn LαLα

٢١٦Chapter 3 -

L+αL+β

α + β

200

Co, wt% Sn20 60 80 1000

300

100

L

α βTE

40

(Pb-SnSystem)

Hypoeutectic & Hypereutectic

Adapted from Fig. 9.8, Callister 7e. (Fig. 9.8 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)

160 µmeutectic micro-constituent


hypereutectic: (illustration only)

β

ββ

ββ

β

Adapted from Fig. 9.17, Callister 7e. (Illustration only)

(Figs. 9.14 and 9.17 from Metals Handbook, 9th ed.,Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)

175 µm

α

α

α

αα

α

hypoeutectic: Co = 50 wt% Sn


T(°C)

61.9eutectic

eutectic: Co =61.9wt% Sn

٢١٧Chapter 3 -

Intermetallic Compounds

Mg2Pb

Note: intermetallic compound forms a line - not an area -because stoichiometry (i.e. composition) is exact.


٢١٨Chapter 3 -

Eutectoid & Peritectic• Eutectic - liquid in equilibrium with two

solidsL α + β

coolheat

intermetallic compound - cementite

coolheat

• Eutectoid - solid phase in equation with two solid phasesS2 S1+S3

γ α + Fe3C (727ºC)

cool

heat

• Peritectic - liquid + solid 1 à solid 2 (Fig 9.21)S1 + L S2

δ + L γ (1493ºC)

٢١٩Chapter 3 -

Eutectoid & Peritectic

Cu-Zn Phase diagram


Eutectoid transition δ γ + ε

Peritectic transition γ + L δ

٢٢٠Chapter 3 -

Iron-Carbon (Fe-C) Phase Diagram• 2 important

points

-Eutectoid (B):γ ⇒ α +Fe3C

-Eutectic (A):L ⇒ γ +Fe3C


Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ+Fe3C

α+Fe3C

α+γ

L+Fe3C

δ

(Fe) Co, wt% C

1148°C

T(°C)

α 727°C = Teutectoid

ASR

4.30Result: Pearlite = alternating layers of α and Fe3C phases

120 µm

(Adapted from Fig. 9.27, Callister 7e.)

γ γγγ

R S

0.76

Ceu

tect

oid

B

Fe3C (cementite-hard)α (ferrite-soft)

٢٢١Chapter 3 -

Hypoeutectoid Steel

Adapted from Figs. 9.24 and 9.29,Callister 7e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ + Fe3C

α+ Fe3C

L+Fe3C

δ

(Fe) Co, wt% C

1148°C

T(°C)

α727°C

(Fe-C System)

C0

0.76

Adapted from Fig. 9.30,Callister 7e.proeutectoid ferritepearlite

100 µm Hypoeutectoidsteel

R S

α

wα =S/(R+S)wFe3C =(1-wα)

wpearlite = wγpearlite

r s

wα =s/(r+s)wγ =(1- wα)

γγ γ

γα

αα

γγγ γ

γ γγγ

٢٢٢Chapter 3 -

Hypereutectoid Steel

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ +Fe3C

α +Fe3C

L+Fe3C

δ

(Fe) Co, wt%C

1148°C

T(°C)

α

Adapted from Figs. 9.24 and 9.32,Callister 7e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

(Fe-C System)

0.76 Co


proeutectoid Fe3C

60 µmHypereutectoid steel

pearlite

R S

wα =S/(R+S)wFe3C =(1-wα)

wpearlite = wγpearlite

sr

wFe3C =r /(r+s)wγ =(1-w Fe3C)

Fe3C

γγγ γ

γγγ γ

γγγ γ

٢٢٣Chapter 3 -

Example: Phase Equilibria

For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following

a) composition of Fe3C and ferrite (α)b) the amount of carbide (cementite) in

grams that forms per 100 g of steelc) the amount of pearlite and

proeutectoid ferrite (α)

٢٢٤Chapter 3 -

Chapter 9 – Phase EquilibriaSolution:

g 3.94g 5.7 CFe

g7.5100 022.07.6022.04.0

100xCFe

CFe

3

CFe3

3

3

=α

=

=−−

=

−−

=α+ α

α

x

CCCCo

b) the amount of carbide (cementite) in grams that forms per 100 g of steel

a) composition of Fe3C and ferrite (α)

CO = 0.40 wt% CCα = 0.022 wt% CCFe C = 6.70 wt% C

3

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

δ

Co , wt% C

1148°C

T(°C)

727°C

CO

R S

CFe C3Cα

٢٢٥Chapter 3 -

Chapter 9 – Phase Equilibriac. the amount of pearlite and proeutectoid ferrite (α)

note: amount of pearlite = amount of γ just above TE

Co = 0.40 wt% CCα = 0.022 wt% CCpearlite = Cγ = 0.76 wt% C

γγ + α

=Co −CαCγ −Cα

x 100 = 51.2 g

pearlite = 51.2 gproeutectoid α = 48.8 g

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

δ

Co , wt% C

1148°C

T(°C)

727°C

CO

R S

CγCα

٢٢٦Chapter 3 -

• Phase diagrams are useful tools to determine:--the number and types of phases,--the wt% of each phase,--and the composition of each phase for a given T and composition of the system.

• Alloying to produce a solid solution usually--increases the tensile strength (TS)--decreases the ductility.

• Binary eutectics and binary eutectoids allow fora range of microstructures.

Summary

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