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MT 30001
Materials Engineering (3-0-0)
Chapter 9 - 1
Instructor: Sumantra MandalDepartment of Metallurgical & Materials Engg
Email: [email protected]
Difference between Crystal & Lattice
Chapter 9 - 2
Difference between Crystal & Lattice
Chapter 9 - 3
Unit Cell
Chapter 9 - 4
Crystal system and Bravais Lattice
Chapter 9 - 5
Cubic Bravais Lattice
Chapter 9 - 6
Lattice Defects
ü Real crystals deviate from the perfect periodicity
üWhile the concept of the perfect lattice is adequate for explaining thestructure-insensitive properties of metal, it is necessary to considera number of types of lattice defects to explain structure-sensitiveproperties
Chapter 9 - 7
Different Types of Lattice Defects
Point defects
Line defects
Surface defects
Chapter 9 - 8
Surface defects
Volume defects
Point Defects
Vacancy
Number of vacant sites (n)
Chapter 9 - 9
Vacancy
Interstitial Impurity
Line Defects - Dislocations
Chapter 9 - 10
Edge dislocation producedby slip in cubic system
Atomic arrangement near edge dislocation
Deformation by slip
Chapter 9 - 11
Schematic drawing of classical idea of slip
Slip in perfect lattice
Shear displacement of plane of atoms over another Relation between shear
stress and displacement
Chapter 9 - 12
Variation of shearing stress with displacement in slip direction
The max. stress at which slip should occur
Slip by dislocation movements
Chapter 9 - 13
• If dislocations don't move, deformation doesn't occur!
Dislocation Motion
• Dislocation moves along slip plane in slip directionperpendicular to dislocation line
• Slip direction same direction as Burgers vector
Edge dislocation
Chapter 9 - 14
Screw dislocation
From Fig. 10.2Callister’s Materials Science and Engineering, Adapted Version.
Critical Resolved Shear Stress for Slip
Slip begins when the shearing stress on the slip plane in the slip direction reaches a threshold value – called as critical resolved shear stress (CRSS)
Chapter 9 - 15
Stress and Dislocation Motion
• Crystals slip due to a resolved shear stress, tR.• Applied tension can produce such a stress.
slip plane
normal, ns
Resolved shear stress: tR =Fs/As
A
tR
Relation between s and tR
tR=FS/AS
Fcos l A/cos f
Applied tensile stress: = F/As
FA
Chapter 9 - 16
AS
tR
FS
Fcos l A/cos f
lF
FS
fnS
AS
AF
fls=t coscosR
• Condition for dislocation motion: CRSS t>tR
• Crystal orientation can makeit easy or hard to move dislocation
10-4 GPa to 10-2 GPa
typically
fls=t coscosR
Critical Resolved Shear Stress
s s s
Chapter 9 - 17t maximum at l = f = 45º
tR = 0l=90°
tR = s/2l =45°f =45°
tR = 0
f=90°
Single Crystal Slip
From Fig. 10.9, Callister’s Materials Science and Engineering, Adapted Version.
Chapter 9 - 18
From Fig. 10.8Callister’s Materials Science and Engineering,Adapted Version.
Ex: Deformation of single crystal
t = s cos l cos f s = 6500 psi
l=35°
f=60°tcrss = 3000 psi
a) Will the single crystal yield? b) If not, what stress is needed?
Chapter 9 - 19
So the applied stress of 6500 psi will not cause the crystal to yield.
s = 6500 psi
t = (6500 psi) (cos35o)(cos60o)
= (6500 psi) (0.41)
t = 2662 psi < tcrss = 3000 psis = 6500 psi
Adapted from Fig. 7.7, Callister 7e.
Ex: Deformation of single crystal
psi 3000t
What stress is necessary (i.e., what is the yield stress, sy)?
)41.0(cos cos psi 3000crss yy s=fls==t
Chapter 9 - 20
psi 732541.0
psi 3000coscos
crss ==fl
t=s\ y
psi 7325=s³s y
So for deformation to occur the applied stress must be greater than or equal to the yield stress
Assignment
Determine the tensile stress that need to be applied along the [1-10] axis of a silver crystal to cause slip on the (1-1-1)[0-11] system. The CRSS is 6 MPa.
Chapter 9 - 21
• Stronger - grain boundariespin deformations
• Slip planes & directions(l, f) change from onecrystal to another.
• t will vary from one
From Fig. 10.10, Callister’s Materials Science and Engineering, Adapted Version.
Slip Motion in Polycrystalss
Chapter 9 - 22
• tR will vary from onecrystal to another.
• The crystal with thelargest tR yields first.
• Other (less favorablyoriented) crystalsyield later.
(Fig. 10.10 is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].)
300 mm
From Fig. 9.1Callister’s Materials Science and Engineering,Adapted Version.
Taxonomy of MetalsMetal Alloys
Steels
Ferrous Nonferrous
Cast Irons Cu Al Mg Ti<1.4wt%C 3-4.5wt%CSteels
<1.4wt%CCast Irons3-4.5wt%C
T(°C) microstructure:
Chapter 9 - 23
From Fig. 7.24Callister’s Materials Science and Engineering, Adapted Version.(Fig. 7.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)
Fe3C cementite
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
gaustenite
g+L
g+Fe3Ca
ferritea+Fe3C
L+Fe3C
d
(Fe) Co , wt% C
Eutectic:
Eutectoid:0.76
4.30
727°C
1148°C
T(°C) microstructure:ferrite, graphite cementite
Iron-Carbon (Fe-C) Phase Diagram
• 2 important points
-Eutectoid (B):g Þ a +Fe3C
-Eutectic (A):L Þ g +Fe3C
T(°C)
C (
cem
entit
e)
1600
1400
1200
1000
L
g (austenite)
g+L
g+Fe3C
L+Fe3C
d
1148°C ASR
g ggg
Chapter 9 - 24
Result: Pearlite = alternating layers of a and Fe3C phases
120 mm
(From Fig. 7.27, Callister Adapted Version.) From Fig. 7.24, Callister Adapted Version.
Fe 3
C (
cem
entit
e)
800
600
4000 1 2 3 4 5 6 6.7
g+Fe3C
a+Fe3C
(Fe) Co, wt% C
a 727°C = Teutectoid
4.30
g ggg
R S
0.76
Ceu
tect
oid
B
Fe3C (cementite-hard)a (ferrite-soft)
Hypoeutectoid Steel
From Figs. 7.24 and 7.29Callister’s Materials Science and Engineering,
C (
cem
entit
e)
1600
1400
1200
1000
800
L
g (austenite)
g+L
g + Fe3C
L+Fe3C
d
1148°C
T(°C)
(Fe-C System)
r sg ga
ggg g
g ggg
Chapter 9 - 25
Science and Engineering, Adapted Version.(Fig. 7.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)
Fe 3
C (
cem
entit
e)
800
600
4000 1 2 3 4 5 6 6.7
a+ Fe3C
(Fe) Co, wt% C
a727°C
C0
0.76
From Fig. 7.30Callister’s Materials Science and Engineering, Adapted Version.
proeutectoid ferritepearlite
100 mm Hypoeutectoidsteel
R S
a
wa =S/(R+S)wFe3
C
=(1-wa)
wpearlite = wgpearlite
r s
wa =s/(r+s)wg =(1- wa)
gg g
ga
aa
Hypereutectoid Steel
C (
cem
entit
e)
1600
1400
1200
1000
800
L
g (austenite)
g+L
g +Fe3C
L+Fe3C
d
1148°C
T(°C)
From Figs. 7.24 and 7.32, Callister’s Materials Science and Engineering,
(Fe-C System)
srFe3C g g
ggg g
ggg g
Chapter 9 - 26
Fe 3
C (
cem
entit
e)
800
600
4000 1 2 3 4 5 6 6.7
a +Fe3C
(Fe) Co, wt%C
a
Science and Engineering, Adapted Version. (Fig. 7.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)
0.76
Co
From Fig. 7.33, Callister’s Materials Science and Engineering, Adapted Version.
proeutectoid Fe3C
60 mmHypereutectoid steel
pearlite
R S
wa =S/(R+S)wFe3C =(1-w a)
wpearlite = wgpearlite
sr
wFe3C =r/(r+s)wg =(1-w Fe3C)
Fe3C
ggg g
Alloying Steel with More Elements
• Teutectoid changes: • Ceutectoid changes: E
utec
toid
(°C
) TiMo Si
W
Cr
eute
ctoi
d(w
t%C
)
NiCr
SiMn
Chapter 9 - 27
From Fig. 7.34Callister’s Materials Science and Engineering, Adapted Version.(Fig. 7.34 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)
From Fig. 7.35Callister’s Materials Science and Engineering, Adapted Version.(Fig. 7.35 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)
TE
utec
toid
wt. % of alloying elements
NiMn
wt. % of alloying elements
Ceu
tect
oid
TiMn
WMo
Steels
Low Alloy High Alloy
low carbon <0.25wt%C
Med carbon0.25-0.6wt%C
high carbon 0.6-1.4wt%C
Additions noneCr,V Ni, Mo
noneCr, Ni Mo
noneCr, V, Mo, W
Cr, Ni, Mo
plain HSLA plainheat
treatableplain tool
austenitic stainless
Name
Chapter 9 - 28Based on data provided in Tables 9.1(b), 9.2(b), 9.3, and 9.4, Callister’s Materials Science and Engineering, Adapted Version.
Uses auto struc. sheet
bridges towers press. vessels
crank shafts bolts hammers blades
pistons gears wear applic.
wear applic.
drills saws dies
high T applic. turbines furnaces V. corros. resistant
Example 1010 4310 1040 4340 1095 4190 304
Additions noneNi, Mo
noneMo
noneMo, W
Cr, Ni, Mo
Hardenability 0 + + ++ ++ +++ 0TS - 0 + ++ + ++ 0EL + + 0 - - -- ++
increasing strength, cost, decreasing ductility
Ferrous Alloys
Iron containing – Steels - cast irons
Nomenclature AISI & SAE10xx Plain Carbon Steels11xx Plain Carbon Steels (resulfurized for machinability) 15xx Mn (10 ~ 20%)
Chapter 9 - 29
40xx Mo (0.20 ~ 0.30%)43xx Ni (1.65 - 2.00%), Cr (0.4 - 0.90%), Mo (0.2 - 0.3%)44xx Mo (0.5%)
where xx is wt% C x 100example: 1060 steel – plain carbon steel with 0.60 wt% C
Stainless Steel -- >11% Cr
Cast Iron
• Ferrous alloys with > 2.1 wt% C
– more commonly 3 - 4.5 wt%C
• low melting (also brittle) so easiest to cast
• Cementite decomposes to ferrite + graphite
Chapter 9 - 30
• Cementite decomposes to ferrite + graphite
Fe3C à 3 Fe (a) + C (graphite)
– generally a slow process
Strategies for Strengthening: 1: Reduce Grain Size
• Grain boundaries arebarriers to slip.
• Barrier "strength"increases withIncreasing angle of
Chapter 9 - 31
Increasing angle of misorientation.
• Smaller grain size:more barriers to slip.
• Hall-Petch Equation: 21 /yoyield dk -+s=s
From Fig. 10.14,Callister’s Materials Science and Engineering, Adapted Version.
(Fig. 10.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)
• Impurity atoms distort the lattice & generate stress.• Stress can produce a barrier to dislocation motion.
4 Strategies for Strengthening: 2: Solid Solutions
• Smaller substitutionalimpurity
• Larger substitutionalimpurity
Chapter 9 - 32
Impurity generates local stress at Aand B that opposes dislocation motion to the right.
A
B
Impurity generates local stress at Cand D that opposes dislocation motion to the right.
C
D
Stress Concentration at Dislocations
Chapter 9 - 33
From Fig. 10.4, Callister’s Materials Science and Engineering,Adapted Version.
Strengthening by Alloying
• small impurities tend to concentrate at dislocations
• reduce mobility of dislocation \ increase strength
Chapter 9 - 34
From Fig. 10.17, Callister’s Materials Science and Engineering,Adapted Version.
Strengthening by alloying
• large impurities concentrate at dislocations on low density side
Chapter 9 - 35
From Fig. 10.18, Callister’s Materials Science and Engineering,Adapted Version.
Ex: Solid SolutionStrengthening in Copper
• Tensile strength & yield strength increase with wt% Ni.
From Fig. 10.16 (a) and (b),Callister’s Materials Science and Engineering,Adapted Version.
Tens
ile s
tren
gth
(MP
a)
300
400
Yie
ld s
tren
gth
(MP
a)
120
180
Chapter 9 - 36
• Empirical relation:
• Alloying increases sy and TS.
21 /y C~s
Adapted Version.
Tens
ile s
tren
gth
(MP
a)
wt.% Ni, (Concentration C)
2000 10 20 30 40 50 Y
ield
str
engt
h (M
Pa)
wt.%Ni, (Concentration C)
600 10 20 30 40 50
• Hard precipitates are difficult to shear.Ex: Ceramics in metals (SiC in Iron or Aluminum).
4 Strategies for Strengthening: 3: Precipitation Strengthening
Large shear stress needed to move dislocation toward precipitate and shear it.
Side View
precipitate
Chapter 9 - 37
• Result:S
~y
1 s
Dislocation “advances” but precipitates act as “pinning” sites with
S.spacing
Top View
Slipped part of slip plane
Unslipped part of slip plane
Sspacing
• Internal wing structure on Boeing 767
Application:Precipitation Strengthening
From chapter-opening photograph, Chapter 11, Callister 5e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)
Chapter 9 - 38
• Aluminum is strengthened with precipitates formedby alloying.
From Fig. 10.31, Callister’s Materials Science and Engineering, Adapted Version.
(Fig. 10.31 is courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)
1.5mm
When the particles are small and/or soft, dislocations can cut and deform the particles as shown in the Figure
Interaction between precipitates and dislocation
Chapter 9 - 39
For the case of averaged non-coherent precipitates Orowan proposed the mechanism illustrated in the Fig. below. The yield stress is determined by the shear stress required to bow a dislocation line between two particles separated by a distance l, where l>R.
Interaction between precipitates and dislocation
Chapter 9 - 40
Schematic drawing, of stages in passage of a dislocation between widely separated obstacles, based on Orowan's mechanism of dispersion hardening
Stage 1: A straight dislocation line approaching two particles.
Stage 2: The line is beginning to bend.
Stage 3: It has reached the critical curvature. The dislocation can then move forward without further decreasing its radius of curvature (R).
and l = 2R, so that the shear stress required to force the dislocation between the obstacles is:
Chapter 9 - 41
between the obstacles is:
Stage 4: Since the segments of dislocation that meet on the other side of the particle are of opposite sign. they can annihilate each other over part of their length, leaving a dislocation loop around each particle
Stage 5: The original dislocation is then free to move on
Chapter 9 - 42
Every dislocation gliding over the slip plane adds one loop around the particle. These loops exert a back stress on dislocation sources which must be overcome for additional slip to take place. This requires an increase in shear stress, with the result that dispersed non-coherent particles cause the matrix to strain-harden rapidly
An alurninum-4% copper alloy has a yield stress of 600 MPa.Estimate the particle spacing in this alloy. Given G ~ 27.6 GPa; b~0.25 nm.
At this strength level we are dealing with a precipitation-hardening alloy that has been aged beyond the maximum strength. The strengthening mechanism is dislocation bypassing of particles.
Exercise
Chapter 9 - 43
G ~ 27.6 GPa; b~0.25 nm; t0= 600/2 = 300 MPa
Stress-Strain Testing• Typical tensile test
machine
specimenextensometer
• Typical tensile specimen
Adapted from Fig. 6.2,Callister 7e.
Chapter 9 - 44
From Fig. 9.9, Callister’s Materials Science and Engineering, Adapted Version.(Fig. 9.9 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.)
gauge length
Linear Elastic Properties
• Modulus of Elasticity, E:(also known as Young's modulus)
• Hooke's Law:
s = E e s
E
F
Chapter 9 - 45
Linear-elastic
E
e
Fsimple tension test
Poisson's ratio, n
• Poisson's ratio, n:eL
e
-n
en = - L
e
metals: n ~ 0.33
Chapter 9 - 46
Units:E: [GPa] or [psi]n: dimensionless
-nmetals: n ~ 0.33ceramics: n ~ 0.25polymers: n ~ 0.40
(at lower temperatures, i.e. T < Tmelt/3)Plastic (Permanent) Deformation
• Simple tension test:
engineering stress, sElastic+Plastic at larger stress
Elastic
Chapter 9 - 47
engineering strain, e
permanent (plastic) after load is removed
ep
plastic strain
Elastic initially
From Fig. 9.16 (a),Callister’s Materials Science and Engineering,Adapted Version.
• Stress at which noticeable plastic deformation hasoccurred.
when ep = 0.002
Yield Strength, sy
sy = yield strengthtensile stress, s
sy
Chapter 9 - 48
Note: for 2 inch sample
e = 0.002 = Dz/z
\ Dz = 0.004 in
From Fig. 9.16 (a),Callister’s Materials Science and EngineeringAdapted Version.
engineering strain, e
ep = 0.002
Tensile Strength, TS
From Fig. 9.17Callister’s Materials Science and Engineering, Adapted Version.
syF = fracture or
ultimate strength
engi
neer
ing
TSst
ress
• Maximum stress on engineering stress-strain curve.
Chapter 9 - 49
• Metals: occurs when noticeable necking starts.• Polymers: occurs when polymer backbone chains are
aligned and about to break.
strain
Typical response of a metal Neck – acts as stress concentrator
engi
neer
ing
stre
ss
engineering strain
• Plastic tensile strain at failure:
Ductility
x 100L
LLEL%
o
of -=
Engineering tensile stress, s
smaller %EL
larger %ELLf
Ao AfLo
Chapter 9 - 50
From Fig. 9.19Callister’s Materials Science and Engineering, Adapted Version..
• Another ductility measure: 100xA
AARA%
o
fo -=
Engineering tensile strain, e
LfAfLo
• Energy to break a unit volume of material• Approximate by the area under the stress-strain
curve.
Toughness
Engineering tensile stress, s
small toughness (ceramics)
large toughness (metals)
Chapter 9 - 51
Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy
very small toughness (unreinforced polymers)
Engineering tensile strain, e
From Fig. 9.19Callister’s Materials Science and Engineering, Adapted Version.
Resilience, Ur
• Ability of a material to store energy
– Energy stored best in elastic region
òe
es= y dUr 0
Chapter 9 - 52
If we assume a linear stress-strain curve this simplifies to
From Fig. 9.21Callister’s Materials Science and Engineering, Adapted Version.
yyr 21
U es@
Elastic Strain Recovery
Chapter 9 - 53
From Fig. 9.23Callister’s Materials Science and Engineering, Adapted Version..
True Stress & StrainNote: specimen area changes when sample stretched
• True stress
• True Strain
iT AF=s
( )oiT llln=e( )( )e+=e
e+s=s1ln
1
T
T
Chapter 9 - 54
From Fig. 9.22Callister’s Materials Science and Engineering, Adapted Version.
Assignment
Chapter 9 - 55
Hardening• An increase in sy due to plastic deformation.
slarge hardening
small hardeningsy0
sy1
Chapter 9 - 56
• Curve fit to the stress-strain response:
sT = K eT( )n“true” stress (F/A) “true” strain: ln(L/Lo)
hardening exponent:n = 0.15 (some steels) to n = 0.5 (some coppers)
e
Assignment
Compute the strain-hardening exponent (n) for an alloy in which a true stress of 415 MPa produces true strain of 0.10. Assume a value of 1035 MPa for K.
Chapter 9 - 57
Instability in Tension
ü Necking generally begins at maximum load during the tensile deformation of a ductile metal.
ü An ideal plastic material in which no strain hardening occurs would become unstable in tension and begin to neck just as soon as yielding took place.
Chapter 9 - 58
soon as yielding took place.
üHowever, a real metal undergoes strain hardening, which tends to increase the load-carrying capacity of the specimen as deformation increases. This effect is opposed by the gradual decrease in the cross-sectional area of the specimen as it elongates.
üNecking or localized deformation begins at maximum load, where the increase in stress due to decrease in the cross-sectional area of the specimen becomes greater than the increase in the load-carrying ability of the metal due to strain hardening.
üThis condition of instability leading to localized deformation
Instability in Tension - Necking
Chapter 9 - 59
üThis condition of instability leading to localized deformation is defined by the condition dP = 0.
P = sAdP = s dA + Ads = 0
This leads to the following relationship,
……….(1)
From the constancy-of-volume relationship,
Instability in Tension - Necking
……….(2)
So that at a point of tensile instability (combining Equation 1 and Equation 2)
Chapter 9 - 60
Graphical interpretation of necking criterion
If the true-stress-true-strain curve is given by the relationship:, where stress is in MPa, what is the ultimate
tensile strength of the material?
Exercise
s = K e( )nhardening exponent:n = 0.15 (some steels)
Hints:
Chapter 9 - 61
( )( )e+=e
e+s=s1ln
1
T
T
sT = K eT( )n“true” stress (F/A) “true” strain: ln(L/Lo)
n = 0.15 (some steels) to n = 0.5 (some coppers)
n=Te ……….How you get the relationship??
Answer: 698 MPa
dtdee =&
00
00 1/)(Lv
dtdL
LdtLLLd
dtd ==
-== ee&
Strain Rate
Strain rate is defined as
Engineering Strain rate
Chapter 9 - 62
00 LdtLdt
Lv
dtdL
LdtL
Ld
dtd T
T ====1)[ln(
0ee&True Strain rate
Here v is crosshead velocity of the machine
A general relationship exists between tensile stress and strain rate, at constant strain and temperature:
where m is known as the strain-rate sensitivity
Relation between Stress and Strain Rate
mTT C )(es &=
Chapter 9 - 63
where m is known as the strain-rate sensitivity
The parameters obtained from tensile tests of a commercially pure aluminum are as follows at a true strain of 0.25.
Determine the change in flow stress for a two order of magnitude change (say 1 to 100 s-1) in strain rate at each of the temperatures.
Exercise
Chapter 9 - 64
change (say 1 to 100 s-1) in strain rate at each of the temperatures.
Variability in Material Properties
• Elastic modulus is material property
• Critical properties depend largely on sample flaws (defects, etc.). Large sample to sample variability.
• Statistics
– Mean xx n
n
S=
Chapter 9 - 65
– Mean
– Standard Deviation( ) 2
1
2
1 úú
û
ù
êê
ë
é
--S
=n
xxs i
n
nx n=
where n is the number of data points
• Design uncertainties mean we do not push the limit.• Factor of safety, N
Ny
working
s=s
Often N isbetween1.2 and 4
• Example: Calculate a diameter, d, to ensure that yield doesnot occur in the 1045 carbon steel rod below. Use a
Design or Safety Factors
Chapter 9 - 66
not occur in the 1045 carbon steel rod below. Use a factor of safety of 5.
( )40002202 /d
N,p
5
Ny
working
s=s 1045 plain
carbon steel: sy = 310 MPa
TS = 565 MPa
F = 220,000N
d
Lo
d = 0.067 m = 6.7 cm
4 Strategies for Strengthening: 4: Cold Work (%CW)
• Room temperature deformation.• Common forming operations change the cross
sectional area:-Forging
A A
force
dieblank
-Rolling
AoAd
roll
Chapter 9 - 67
Adapted from Fig. 11.8, Callister 7e.
Ao Adblank
force-Drawing
tensile force
AoAddie
die
-Extrusion
ram billet
container
containerforce
die holder
die
Ao
Adextrusion
100 x %o
doA
AACW
-=
roll
Ao
• Ti alloy after cold working:
• Dislocations entanglewith one anotherduring cold work.
• Dislocation motionbecomes more difficult.
Dislocations During Cold Work
Chapter 9 - 68
becomes more difficult.
From Fig. 5.10, Callister’s Materials Science and Engineering, Adapted Version.
(Fig. 5.10 is courtesy of M.R. Plichta, Michigan Technological University.)
0.9 mm
Result of Cold Work
Dislocation density =
– Carefully grown single crystal
à ca. 103 mm-2
– Deforming sample increases density
à 109-1010 mm-2
total dislocation lengthunit volume
Chapter 9 - 69
à 109-1010 mm-2
– Heat treatment reduces density
à 105-106 mm-2
• Yield stress increasesas rd increases: large hardening
small hardening
s
e
sy0sy1
Effects of Stress at Dislocations
From Fig. 10.5, Callister’s Materials Science and Engineering,Adapted Version.
Chapter 9 - 70
Impact of Cold Work
• Yield strength (sy) increases.• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.
As cold work is increased
Chapter 9 - 71
From Fig. 10.20Callister’s Materials Science and Engineering, Adapted Version.
• What is the tensile strength &ductility after cold working?
%6.35100 x %2
22=
p
p-p=
o
do
r
rrCW
Cold Work Analysis
700
yield strength (MPa) tensile strength (MPa)
800
ductility (%EL)60
Do =15.2mm
Cold Work
Dd =12.2mm
Copper
Chapter 9 - 72
From Fig. 10.19, Callister’s Materials Science and Engineering, Adapted Version.(Fig. 10.19 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.)
% Cold Work
100
300
500
700
Cu
200 40 60
sy = 300MPa
300MPa
% Cold Work
200Cu
0
400
600
20 40 60% Cold Work
20
40
20 40 6000
Cu340MPa
TS = 340MPa
7%
%EL = 7%
• Results forpolycrystalline iron:
From Fig. 9.20 Callister’s Materials Science and Engineering,Adapted Version.
s-e Behavior vs. Temperature-200°C
-100°C
25°C
800
600
400
200
0
Strain
Str
ess
(MP
a)
0 0.1 0.2 0.3 0.4 0.5
Chapter 9 - 73
• sy and TS decrease with increasing test temperature.• %EL increases with increasing test temperature.• Why? Vacancies
help dislocationsmove past obstacles.
2. vacancies replace atoms on the disl. half plane
3. disl. glides past obstacle
Strain
1. disl. trapped by obstacle
obstacle
q Annealing of the cold worked structure at high temperature softens the metal and reverts to a strain-free condition.
q Annealing restores the ductility to a metal that has been severely strain hardened.
q Annealing can be divided into three distinct processes:
Effect of Heating (Annealing) After Cold Working
Chapter 9 - 74
q Annealing can be divided into three distinct processes:
ü Recovery
ü Recrystallization
ü Grain growth
Annihilation reduces dislocation density.
Recovery
• Scenario 1Results from diffusion
Dislocations annihilate and form a perfect atomic plane.
extra half-plane of atoms
extra half-plane
atoms diffuse to regions of tension
Chapter 9 - 75
• Scenario 2
4. opposite dislocations meet and annihilate
extra half-plane of atoms
2. grey atoms leave by vacancy diffusion allowing disl. to “climb”
tR
1. dislocation blocked; can’t move to the right
Obstacle dislocation
3. “Climbed” disl. can now move on new slip plane
Recovery
Various stages in the
Chapter 9 - 76
Various stages in the recovery of a plastically deformed material
Recrystallization
Chapter 9 - 77
(a)
Schematic illustration of nucleation event: (a) subgrain (SG) initially grows within Grain I and reaches the critical size which allows it to overcome the capillary force; (b) subsequently it can bulge into Grain II as a new strain free grain
(b)
• New grains are formed that:-- have a small dislocation density-- are small-- consume cold-worked grains.
From Fig. 10.21
0.6 mm 0.6 mm
Recrystallization
Chapter 9 - 78
From Fig. 10.21 (a),(b),Callister’s Materials Science and Engineering, Adapted Version.(Fig. 10.21 (a),(b) are courtesy of J.E. Burke, General Electric Company.)33% cold
workedbrass
New crystalsnucleate after3 sec. at 580°C.
• All cold-worked grains are consumed.
From Fig. 10.21 (c),(d),
0.6 mm0.6 mm
Further Recrystallization
Chapter 9 - 79
(c),(d), Callister’s Materials Science and Engineering, Adapted Version.(Fig. 10.21 (c),(d) are courtesy of J.E. Burke, General Electric Company.)After 4
secondsAfter 8seconds
• At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy)
is reduced.0.6 mm 0.6 mm
From Fig. 10.21 (d),(e)Callister’s Materials Science and Engineering, Adapted Version.(Fig. 10.21 (d),(e) are courtesy of J.E. Burke, General Electric
Grain Growth
Chapter 9 - 80
After 8 s,580ºC
After 15 min,580ºC
J.E. Burke, General Electric Company.)
• Empirical Relation:
Ktdd no
n =-elapsed time
coefficient dependenton material and T.
grain diam.at time t.
exponent typ. ~ 2
TR
From Fig. 10.22,
º
TR = recrystallization temperature
Chapter 9 - 81
From Fig. 10.22, Callister’s Materials Science and Engineering,Adapted Version.
º
Recrystallization Temperature, TR
TR = recrystallization temperature = point of highest rate of property change1. Tm => TR » 0.3-0.6 Tm (K)2. Due to diffusion à annealing timeà TR = f(t)
shorter annealing time => higher TR
Chapter 9 - 82
3. Higher %CW => lower TR
4. Pure metals lower TR due to dislocation movements
• Dislocation can move easily in pure metals => lower TR
• Recovery : The restoration of the physical properties of thecold worked metal without any observable change inmicrostructure. Strength is not affected.
• Recrystallization : The cold worked structure is replaced bya new set of strain-free grains due to migration of high angle grain
Summary
Chapter 9 - 83
a new set of strain-free grains due to migration of high angle grainboundaries. Hardness and strength decrease but ductilityincreases.
• Grain growth : Occurs at higher temperature where some ofthe recrystallized fine grains start to grow rapidly.
