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FE ReviewMaterials Properties
Jeffrey W. FergusMaterials EngineeringOffice: 284 Wilmore
Phone: 844-3405email: [email protected]
Electrical Properties
• Electrical resistance– resistance (R) = resistivity (ρ) length (l) / area (A)– resistivity is a material property– conductivity (σ) = 1 / resistivity (ρ)
l
A– conductivity (σ) = 1 / resistivity (ρ)
• Temperature dependence – with increasing temperature…– metals: resistance increases (conductivity decreases)– semiconductors: conductivity increases (resistivity decreases)
• extrinsic: like metals in intermediate temperatures
– insulators: conductivity increases (resistivity decreases)
A
Mechanical Properties
• Stress-strain relationships– engineering stress and strain– stress-strain curve
• Testing methods• Testing methods– tensile test– endurance test– impact test
Tensile Test
thickness
Control length (l)
Measure force (F) with load cell
oo
ol
ll
ll ∆=−=ε=Strain
lengthwidth
Measure force (F) with load cell
twF
AF
⋅==σ=Stress
Reduced section used to limit portion of sample undergoing deformation
Stress-Strain Curve
Ultimate Tensile Strength
Yield Point
Elastic Limit
Force decreases due to neckingS
tres
s
Strain
Slope = E (Young’s Modulus)
Percent Elongation(total plastic deformation)
Elastic Limit
Proportionality Limit
True/Engineering Stress/Strain
oE A
F=σoo
oE l
ll
ll ∆=−=εEngineering
(initial dimensions)
Stress Strain
iT A
F=σ ∫
==ε
i
o
l
l o
iT l
lln
ldl
( )1+ε=ε ET ln
( )EET ε+σ=σ 1ooii lAlA ⋅=⋅
True(instantaneous
dimensions)
Usingand
True/Engineering Stress/Strain
True
True stress does not decrease
Str
ess
Strain
Engineering
Decrease in engineering stress due to decreased load required in the reduces cross-sectional area of the neck.
Strain Hardening
Str
ess
Plastic deformation require larger load after deformation.
Str
ess
Strain
Onset of plastic deformation after
reloading
after deformation.Sample dimensions are decreased, so stress is even higher
Bending Test
Four-point Three-point
h
wFF/2F/2
LL/2
By summing moment in cantilever beam
22
3
wh
FLmax =σ
Tension at bottom, compression at top
Hardness
• Resistance to plastic deformation• Related to yield strength• Most common indentation test
– make indentation– make indentation– measure size or depth of indentation– macro- and micro- tests
• Scales: Rockwell, Brinell, Vickers, Knoop
Impact
Toughness: combination of strength and ductility -energy for fracture
hi
hf
Fracture energy = mghi -mghf
Charpy V-notch
Ductile-Brittle Failure
• Ductile– plastic deformation– cup-cone / fibrous
fracture surface
Ductile-Brittle Transition Temperature
(DBTT)• Brittle
– little or no plastic deformation
– cleaved fracture surface
Temperature
Fra
ctur
e E
nerg
y
(DBTT)
Creep / Stress Relaxation
• Load below yield strength - elastic deformation only• Over long time plastic deformation occurs• Requires diffusion, so usually a high-temperature
processprocess• Activation energy, Q (or EA)
−⋅=
−⋅=ε=kTE
expART
QexpA A
ɺratecreep
Creep /Stress Relaxation
CreepFF Stress Relaxation
FF
time
fixed strain
fixed load
time
Permanent deformation
Fatigue
Fatigue Limit(ferrous metals)
Repeated application of load - number of cycles, rather than time important.
σmax
0
Number of Cycles to Failure
Str
ess (ferrous metals)
σmin
σave ∆σ
σmax
σmin
0
Corrosion Resistance
• Thermodynamics vs. Kinetics– thermodynamics - stable phases– kinetic - rate to form stable phases
• Active vs. Passive– active: reaction products ions or gas - non protective– passive: reaction products - protective layer
• Corrosion resistance– inert (noble): gold, platinum– passivation: aluminum oxide (alumina) on aluminum,
chromia on stainless steel
Electrode Potential
• Tendency of metal to give up electron• Oxidation (anode)
– M = M2+ + 2e- (loss electrons)
• Reduction (cathode)• Reduction (cathode)– M2+ + 2e- = M (gain electrons)
• LEO (loss electrons oxidation) goes GER (gain electrons reduction)
Corrosion Reactions• Oxidation - metal (anode)
– M = M2+ + 2e-
• Reduction - in solution (cathode)– 2H+ + 2e- = H2
– 2H+ + ½O2 + 2e- = H2O– H O + ½O + 2e- = 2OH-– H2O + ½O2 + 2e- = 2OH-
• Overall Reactions– M + 2H+ =M2+ + H2
– M + 2H+ + ½O2 = M2+ + H2O– M + H2O + ½O2 = M2+ + 2OH- = M(OH)2
Electromotive Force
• Gibbs Free Energy (∆G) =-nFE (Electromotive Force)– n = number of electrons, F = Faraday’s Constant– favorable: energy decrease (-) = positive voltage
• Fe2+ + 2e- = Fe: Ered = +0.440 V• Fe2+ + 2e- = Fe: Ered = +0.440 V• Fe = Fe2+ + 2e-: Eox = -0.440 V• H2O = 2H+ + ½O2 +2e-: Ered = +1.229 V• Fe + 2H+ + ½O2 = Fe2+ + H2O: E = 0.789 V
– E does not change with number of moles (∆G does)– E must be corrected for non-standard state
• concentration of H+ (i.e. pH), oxygen pressure…
Galvanic Corrosion / Protection
• At joint between dissimilar metals– reaction rate of active metal increases– reaction of less active metal decreases
• Galvanic corrosion– high corrosion rate at galvanic couple
• presence of Cu increase the local corrosion rate of Fe
• Galvanic protection– galvanized steel
• presence of Zn decreases the local corrosion rate of Fe
– galvanic protection• Mg or Zn connected to Fe decrease corrosion rate
Zn
Fe
Fe Cu
Waterline Corrosion
• Oxygen concentration in water leads to variation in local corrosion rates
Higher corrosion rate near oxygen accessoxygen access
Rings of rust left from water drops
Rust just below water surface
Materials Processing
• Diffusion• Phase Diagrams
– Gibb’s phase rule– lever rule– lever rule– eutectic system / microconstituents– Fe-Fe3C diagram (ferrous metals)
• Thermal-mechanical processing
Diffusion
• Atoms moving within solid state• Required defects (e.g. vacancies)• Diffusion thermally activated• Diffusion constant follows Arrhenius relationship
−=
−=kTE
expDRT
QexpDD A
oo
Gas constant
Boltzman’s constant Temperature
Activation Energy
Steady-State Diffusion
• Fick’s first law (1-D)• J = flux (amount/area/time)• For steady state
∂∂−=
xC
DJ
∆∆−=
xC
DJ ∆x
∆C
∆x
sm
massm
m
mass
sm
J2
32=
−=
Phase Equilibria
• Gibb’s Phase Rule• P + F = C + 2 (Police Force = Cops + 2)
– P = number of phases– F = degrees of freedom– C = number of components (undivided units)– 2: Temperature and Pressure– 2: Temperature and Pressure
• One-component system– F = 1 + 2 - P = 3 - P
• Two-component system– F = 2 + 2 - P = 4 - P
• Two-component system at constant pressure– F = 2 + 1 - P = 3 - P
“2” becomes “1” at constant pressure
Pressure-Temperature Diagram
Single-phase area: can change T
Two-phase line: Change T (P) require specific change in P (T)
(F=1)
One component: H2OIf formation of H2 and O2 were considered there would be two
components (H and O)
water
ice
watervapor
Temperature
Pre
ssur
e
Single-phase area: can change T and P independently
(F=2)
Three-phase point: One occurs at specific T and P (triple point)
(F=0)
Phase Diagrams
L
δ
δ + γ δ + L
Two-component @ constant pressureThree-phase - horizontal line
Eutectic
PeritecticL +solid (δ) → solid (γ)
α
γ
β (pure B, negligible
solubility of A)
α + γ
γ + β
α + β
γ + L β + L
Composition (%B)
Tem
pera
ture Eutectic
L → 2 solids (γ + β)
Eutectoidsolid (γ) → 2 solids (α + β)
BA
Lever Law• Phase diagram give compositions of phases
– two-phase boundaries in 2-phase mixture
• Mass balance generate lever law
LiquidComp.
(XL)
SolidComp.
(XS)
AlloyComp.(Xalloy)
Opposite arm over total length
Right arm for solid
L
S
Composition (%B)A B
Tem
pera
ture
(XL)(XS) (Xalloy)
SL
alloyL
XX
XXsolid%
−−
=
SL
Salloy
XX
XXliquid%
−−
=
Right arm for solid
Left arm for liquid
70 wt% Pb - 30 wt% Sn
256°C
First solid
( ) %27)(%3.18.)(%8.61)(%3.18)(%30
%8.61.% =−−=
PbSnliqSnPbSnalloySn
Snliq
At 183.1°C
(Pb)L
12.8 wt% Sn
( ) %73)(%3.18.)(%8.61)(%30.)(%8.61
%3.18.% =−−=
PbSneutSnalloySneutSn
SnPbprim
(Pb)L
70 wt% Pb - 30 wt% Sn
256°C
First solid
( ) %15)(%3.18.)(%8.97)(%3.18)(%30
%8.97% =−−=
PbSnliqSnPbSnalloySn
Snβ
At 182.9°C
(Pb)
12.8 wt% Sn
( ) %85)(%3.18.)(%8.97)(%30.)(%8.97
%3.18% =−−=−
PbSnliqSnalloySnliqSn
SnphasePb
(Pb)
Eutectic(Pb)+β
Microconstituents
( ) )(%3.18)(%30 =−= PbSnalloySn
( ) %73)(%3.18.)(%8.61)(%30.)(%8.61
%3.18.Prim% =−−=
PbSneutSnalloySneutSn
SnPb
Eutectic Microsconstituent ((Pb)+βSn)
Primary Pb
( ) %27)(%3.18.)(%8.61)(%3.18)(%30
%8.61% =−−=
PbSnliqSnPbSnalloySn
SnL
( ) %55)(%3.18.)(%8.97)(%3.18.)(%8.61
%8.97% . =−−=
PbSnliqSnPbSneutSn
Sneutinβ
( ) %45)(%3.18.)(%8.97.)(%8.61.)(%8.97
%3.18% . =−
−=PbSnliqSneutSnliqSn
SnPb eutin
Phases in Eutectic Microsconstituent
Phases in Microconstituents
(Pb) 73 g primary (Pb)
27 g eutecticβ
12 g (Pb) (45% of 27 g)
(Pb) + βSn15 g βSn (55% of 27 g)
Total amounts in 100 g sampleTotal (Pb) = 73 + 22 = 85 g Total βSn = 15 g(same as directly from the lever law)
Fe-Fe3C Phase Diagram
Austenite Cementite
Ferrite
Pearlite (ferrite + cementite)%C = 0.77%
HypereutectoidHypoeutectoid
Cast Irons
Steels
Time-Temperature-Transformation (TTT) Diagram
fs727°C
800°C
Decomposition of Austenite at fixed temperature Pearlite: High Temp
slow nucleation
Coarse pearliteps
bs
ms
mf
pf
bf
200°C
100°C
KeyMain symbolf = ferritep = pearliteb = bainitec = cementite
(Fe3C)Subscriptss = startf = finish
Tem
per
atu
re
Log TimeMartensite
athermal (diffusionless)
Bainite: Diffusion slow
for pearlite
Coarse pearlite
Fine pearlite
Quench / Hardenability / Tempering
• Quench - rapidly cool– in steel: cool fast enough to Ms to prevent pearlite / bainite
formation
• Hardenability• Hardenability– ease of forming martensite in steels– alloying elements inhibit pearlite / bainite formation, promote
martensite formation
• Tempering of steels– reheating martensite to form transition carbides– improve toughness
Cold Working• Plastic deformation creates dislocations, which
increases strength / decreases ductility• Reduction in Area used to quantify degree of cold
working%
AAfA
RA%CW%i
i 100⋅−==
lwlw ⋅−⋅%
lwlwlw
RA%ii
ffii 100⋅⋅
⋅−⋅=
%l
llRA%
i
fi 100⋅−=
if wwfor ≅
%d
dd%
d
dd
RA%
i
fi
i
fi
100100
4
442
22
2
22
⋅−
=⋅⋅π
⋅π−
⋅π
=
Cold Worked Properties
400
500
600
Str
ess
(MP
a) 10
12
14
16
Percen
t Elo
ng
ation
Yield StrengthTensile Strength
0
100
200
300
0 10 20 30 40 50 60 70 80
Percent Cold Work
Str
ess
(MP
a)
0
2
4
6
8
Percen
t Elo
ng
ation
Tensile StrengthPercent Elongation
Balancing Strength / Ductility
300
400
500
600
Str
ess
(MP
a)
20
25
30
35
Percen
t Elo
ng
ation
Yield Strength
Tensile Strength
Percent Elongation
0
100
200
0 10 20 30 40 50 60 70 80
Percent Cold Work
Str
ess
(MP
a)
0
5
10
15
Percen
t Elo
ng
ation
Sy > 310 MParequires
%CW > 22%Elongation > 10%
requires%CW < 31%
Both Propertiesrequires
22% < %CW < 31%
Balancing Strength / Toughness
400
450
500
550
600
Str
ess
(MP
a)30
35
40
45
50Yield Strength
Tensile Strength
Fracture Toughness Fractu
re To
ug
hn
ess (K
σy = 250 MPa13% CW
Sy > 250 MPaand
KIc > 16 MPa m½
requires13% < %CW < 39%
100
150
200
250
300
350
0 10 20 30 40 50 60 70Percent Cold Work
Str
ess
(MP
a)
0
5
10
15
20
25
Fractu
re To
ug
hn
ess (KIc ) (M
Pa m
0.5)
KIc = 16 MPa m0.5
39% CW
31% CWSy = 364 MPa
KIc = 22 MPa m0.5
13% < %CW < 39%
Examplefor 31% CW
Sy = 364 MPaKIc = 22 MPa m½
Cold Work / Anneal / Hot Work• Annealing can eliminate effect of cold work
– recovery - stress relief, little change in properties– recrystallization - elimination of dislocations, decrease in
strength, increase in ductility– grain growth - increase in grain size, decreases both
strength and ductility
• Hot working• Hot working– deforming at high enough temperature for immediate
recrystallization– list cold-working and annealing at the same time– no increase in strength– used for large deformation– poor surface finish - oxidation– after hot working, cold working used to increase strength and
improve surface finish
Organization from 1996-7 Review Manual(same topics in 2004 review manual)
• Crystallography• Materials Testing• Metallurgy
Crystallography
• Crystal structure– atoms/unit cell– packing factor– coordination number– coordination number
• Atomic bonding• Radioactive decay
Bravais Lattice
Crystal System Centering
P: Primitive: (x,y,z)
(x,y,z): Fractional coordinates -proportion of axis length, not absolute distanct
a
b
c
β α
γ
P: Primitive: (x,y,z)
I: Body-centered: (x,y,z); (x+½,y+½,z+½)
C: Base-centered: (x,y,z); (x+½,y+½,z)
F: Face-centered: (x,y,z); (x+½,y+½,z)(x+½,y,z+½); (x,y+½,z+½)
Centering must apply to all atoms in unit cell.
Bravais Lattices (14)
Crystal System Parameters Primitive
(Simple) Body-
Centered Face-
Centered Base-
Centered
Cubic a=b=c
α=β=γ=90° X X X
Tetragonal a=b≠c
α=β=γ=90° X X
α=β=γ=90°
Orthorhombic a≠b≠c
α=β=γ=90° X X X X
Rhombohedral a=b=c
α=β=γ≠90° X
Hexagonal a=b≠c
α=β=90°, γ=120° X
Monoclinic a≠b≠c
α=γ=90°, β≠120° X X
Triclinic a≠b≠c
α≠β≠γ≠90° X
Atoms Per Unit Cell
• Corners - shared by eight unit cells (x 1/8)– (0,0,0)=(1,0,0)=(0,1,0)=(0,0,1)=(1,1,0)
=(1,0,1)=(0,1,1)=(1,1,1)
• Edges - shared by four unit cells • Edges - shared by four unit cells (x 1/4)– (0,0,½)= (1,0,½)= (0,1,½)= (1,1,½)
• Faces - shared by two unit cells (x 1/2)– (½,½,0)= (½,½,1)
Common Metal Structures
• Face-Centered Cubic (FCC)– 8 corners x 1/8 + 6 faces x 1/2 – 1 + 3 = 4 atoms/u.c.
• Body-Centered Cubic (BCC)– 8 corners x 1/8 + 1 center – 8 corners x 1/8 + 1 center – 1 + 1 = 2 atoms/u.c.
• Hexagonal Close-Packed (HCP)– 8 corners x 1/8 + 1 middle – 1 + 1 = 2 atoms/u.c.– 12 hex. Corner x 1/6 +2 face x 1/2 + 3
middle = 6 atoms/u.c.
Packing Factor
• Fraction of space occupied by atoms• For FCC
cba
rFP i
⋅⋅= ∑
334 π
..
ar
raraa2
44diagonalface 22 =⇒⋅=+=
( ) ( )3434
• For BCCa
r ( ) ( )740
23
24
443
334
3
334
.r
a
r.F.P =π=
π⋅=
π⋅=
r3
4ar4aaadiagonalbody 222 =⇒⋅=++=
( ) ( )680
83
34
223
334
3
334
.r
a
r.F.P =π=
π⋅=
π⋅=
Density
volumemass
.c.uvolume
moleatom
molemass
.c.uatom
Density =
⋅
⋅
=
For nickel:- Atomic weight = 58.71 g/mole
( ) 338249158
1052393106020
71584
cm
g.
cmx.moleatom
x.
moleg
..c.u
atom
Density =⋅
⋅
=−
- Atomic weight = 58.71 g/mole- Lattice parameter = 3.5239 Å=3.5239 x 10-8 cm- Avogadro’s No. = 6.02 x 1023 = 0.602 x 1024 = atoms/mole
Octahedral Site (CN=6)
In FCC:- Center (½,½,½)- Edges (0,0,½),(0,½,0),(½,0,0)- 4 per unit cell- All filled - NaCl structure
8-sided shape
Tetrahedral Site (CN=4)In FCC:- Divide cell into 8 boxes - center of small box- (¼,¼,¼),(¾,¼,¼),(¼,¾,¼),(¾,¾,¼)(¼,¼, ¾)(¾,¼, ¾),(¼,¾, ¾)(¾,¾, ¾)-8 per unit cell-All filled - CaF2 structure; half-filled - ZnS-All filled - CaF2 structure; half-filled - ZnS
4-sided shape
Radius Ratio Rules
Critical Radius for CN 8 = 0.732CN 8
Critical radius is size of atom which just fits in siteDefine minimum for bonding (i.e. atoms must touch to bond)
Critical Radius for CN 8 = 0.732
Critical Radius for CN 6 = 0.414
Critical Radius for CN 4 = 0.225
CN 6
CN 4
CN 3planar
Close Packed Plane
A BA BA C
HCP: ABABABABABABABABFCC: ABCABCABCABCABCSame packing density (0.74)Same coordination (CN=12)
Miller Indices
Planes
Directions
(hkl)
{hkl}
[hkl]
<hkl>
specific
family
specific
family
- No commas- No fractions- Negative indicated by bar over number<hkl> family
A family of planes includes all planes which are equivalent by symmetry - depends on crystal system.- For cubic: (110),(011) and (101) are all {110}- For tetragonal: (011) and (101) are {101}
but (110) is not (c≠a)
over number
Miller Indices - Directions
ba
c
-1/3
1/2-1
x1/2
y-1
z-1/3 (x 6)
[ ]263
1
1/4
1/2
x1
y1/4
z1/2 (x 4)
[ ]214
[ ]263
Miller Indices - Planes
31
21
ba
c
41
31
21
ba
c
41
x1/44
y-1/3-3
z-1/2-2
( )234
interceptreciprocal
Atomic Bonding
• Covalent– sharing electrons– strong– directional
• Metallic– metal ions in sea or
electrons– moderately strong– directional
• Ionic– trading of electrons– electrostatic attraction or
ions– strong– non-directional
– non-directional
• Secondary– Van der Waals– H-bonding– electrostatic attraction of
electric dipole (local charge distribution
– weak
Radioactive Decay
• Loss of electrons/protons/neutrons– alpha - 2 protons / two neutrons (i.e He nucleus)– beta - electrons– gamma - energy
• Exponential decay
−= t
expNNtime
• Exponential decay
τ
−= texpNN o
( ) ( )22 2
121
21
21
ln
tln
N
Nln
texpNNN
o
ooo −=τ⇒=
−=
τ⇒==
( )
⋅−=
⋅−=21
21
69302t
t.expN
tlnt
expNN oo half life
amount
original amount
time constant