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CHAPTER 7: DISLOCATIONS AND STRENGTHENING. In Ch.6 Plastic (permanent) & Elastic (reversible) Yield Strength and hardness are a measure of materials resistance to deformation In microscopic scale what is going on ? Net movement of large number of atoms in response to an applied stress - PowerPoint PPT Presentation
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Chapter 7-
CHAPTER 7: DISLOCATIONS AND STRENGTHENING
• In Ch.6– Plastic (permanent) & Elastic (reversible)
• Yield Strength and hardness are a measure of materials resistance to deformation
– In microscopic scale what is going on ?• Net movement of large number of atoms in response to
an applied stress• Involves movement of dislocations, in Ch. 4
– How does strengthening happen ?– Why do some materials are stronger than others ?– How can one manipulate dislocations and their
motion ?
Chapter 7 -
Dislocations & Materials Classes
• Covalent Ceramics (Si, diamond): Motion hard. -directional (angular) bonding
• Ionic Ceramics (NaCl): Motion hard. -need to avoid ++ and - - neighbors.
+ + + +
+++
+ + + +
- - -
----
- - -
• Metals: Disl. motion easier. -non-directional bonding -close-packed directions for slip. electron cloud ion cores
++
++
++++++++ + + + + +
+++++++
Chapter 7 -
Dislocation Motion
Dislocations & plastic deformation• Cubic & hexagonal metals - plastic deformation by
plastic shear or slip where one plane of atoms slides over adjacent plane by defect motion (dislocations).
• If dislocations don't move, deformation doesn't occur!
Adapted from Fig. 7.1, Callister 7e.
Chapter 7- 3
• Produces plastic deformation !• Bonds are incrementally broken and re-formed. Much less force is needed , why ?
Plasticallystretchedzincsinglecrystal.
• If dislocations don't move, plastic deformation doesn't happen!
Adapted from Fig. 7.1, Callister 6e. (Fig. 7.1 is adapted from A.G. Guy, Essentials of Materials Science, McGraw-Hill Book Company,New York, 1976. p. 153.)
Adapted from Fig. 7.9, Callister 6e. (Fig. 7.9 is from C.F. Elam, The Distortion of Metal Crystals, Oxford University Press, London, 1935.)
Adapted from Fig. 7.8, Callister 6e.
DISLOCATION MOTION
SEE THE MOVIE AGAIN !
How do we generate the dislocation motion ?
SLIP PLANE
Chapter 7-
DISLOCATION MOTION
t0t1t2t3
The motion of a single dislocation across the plane causesthe top half of the crystal to move (to slip) with respect tothe bottom half but we can not have to break all the bondsacross the middle plane simultaneously (which wouldrequire a very large force).
The slip plane (O0-O1)– the crystallographic plane of dislocationmotion.
O1 O0
Initial statedeformed apply force
Chapter 7-
STRESS and Dislocation Motion
NOT be parallel to the dislocation line !
Dislocation line
Chapter 7 -
Dislocation Motion
• Dislocation moves along slip plane in slip direction perpendicular to dislocation line
• Slip direction same direction as Burgers vector
Edge dislocation
Screw dislocation
Adapted from Fig. 7.2, Callister 7e.
Chapter 7-
STRESS AND DISLOCATION MOTION
Chapter 7-
STRESS field around dislocations
WHY is there a stress field ?Atoms try to relax by trying to achieve their positions for the case if there was not a dislocation in the vicinity !
Bonds are Stretched
Bonds are Compressed
Chapter 7-
Stress fields of dislocations interacting !The strain fields around dislocations interact with each other. Hence, they exert force on each other.
Edge dislocations, when they are in the same plane, they repel each other if they have the samesign (direction of the Burgers vector). WHY ?
They can attract and annihilate if they have opposite signs. PROVE !
Chapter 7-
Dislocation Density
• The number of dislocations in a material is expressed using the term dislocation density - the total dislocation length per unit volume or the # of dislocations intersecting a unit area. Units are mm / mm3, or just / mm
• Dislocation densities can vary from 103 mm-2 in carefully solidified metal crystals to 1010 mm-2 in heavily deformed metals.
Where do Dislocations come from, what are their sources ?• Most crystalline materials, especially metals, have dislocations
in their as-formed state, mainly as a result of stresses (mechanical, thermal...) associated with the manufacturing processes used.
• The number of dislocations increases dramatically during plastic deformation.
• Dislocations spawn from existing dislocations, grain boundaries and surfaces and other “defects” .
NICE SIMULATIONS => http://www.ims.uconn.edu/centers/simul/movie/
Chapter 7 -
Slip System– Slip plane - plane allowing easiest slippage
• Wide interplanar spacings - highest planar densities
– Slip direction - direction of movement - Highest linear densities
– FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed)
=> total of 12 slip systems in FCC– in BCC & HCP other slip systems occur
Deformation Mechanisms
Adapted from Fig. 7.6, Callister 7e.
Chapter 7-
SLIP SYSTEMS !!!
• Dislocations move with ease on certain crystallographic planes and along certain directions on these planes ! – The plane is called a slip plane– The direction is called a slip direction – Combination of the plane of slip and direction is a
slip system• The slip planes and directions are those of highest
packing density. – The distance between atoms is shorter than the
average…High number of coordination along the planes also important !
PLS SEE TABLE 7.1 in ur books , page 180 !!!
Chapter 7-
Concept Check
Page 181, in 7th edition !
Chapter 7 -
Stress and Dislocation Motion• Crystals slip due to a resolved shear stress, R. • Applied tension can produce such a stress.
slip plane
normal, ns
Resolved shear stress: R =Fs/As
slip
directi
on
AS
R
R
FS
slip
directi
on
Relation between and R
R =FS /AS
Fcos A/cos
F
FS
nS
AS
A
Applied tensile stress: = F/A
slip
directi
on
FA
F
coscosR +
Chapter 7 -
• Condition for dislocation motion: CRSS R
• Crystal orientation can make it easy or hard to move dislocation
10-4 GPa to 10-2 GPa
typically
coscosR
Critical Resolved Shear Stress
maximum at = = 45º
R = 0
=90°
R = /2=45°=45°
R = 0
=90°
Chapter 7-
Resolving the Applied Stress on a SLIP PLANE !
• Shear Stress has to be resolved on to the slip planes as;– Shear Stress is needed for dislocations to move / slip– Dislocations can only move on slip planes, and these
planes are rarely on axis with the applied force !
We should resolve the force applied in a tensile test, F, on to the cross-sectional area A where the slip is going to take place;
Chapter 7-
Critical Resolved Shear Stress => Slip in Single X’tals
Macroscopically;Q: when do materials plastically deform / yield ?
Stress > YSQ: How does plastic deformation take place ?
Dislocation Motion ( Elastic deformation ??)Q: On what planes does dislocations move ?
Slip PlanesQ: At stress = YS, what would be the minimum resolved shear stress needed to act on dislocations to initiate dislocation motion, onset of yield/plastic deformation ?
HIMMM ! Lets think !!!
Chapter 7-
The Critical Resolved Shear StressThe minimum shear stress required to initiate slip is termed:the critical resolved shear stress
Chapter 7 -
Single Crystal Slip
Adapted from Fig. 7.8, Callister 7e.
Adapted from Fig. 7.9, Callister 7e.
Plasticallystretched
zincsingle
crystal.
A large number of dislocations are generated on slip planes, as they leave the system they form these “shear bands”
Chapter 7 -
Ex: Deformation of single crystal
So the applied stress of 6500 psi will not cause the crystal to yield.
cos cos 6500 psi
=35°
=60°
(6500 psi) (cos35)(cos60)
(6500 psi) (0.41)
2662 psi crss 3000 psi
crss = 3000 psi
a) Will the single crystal yield? b) If not, what stress is needed?
= 6500 psi
Adapted from Fig. 7.7, Callister 7e.
Chapter 7 -
Ex: Deformation of single crystal
psi 732541.0
psi 3000
coscoscrss
y
What stress is necessary (i.e., what is the yield stress, y)?
)41.0(cos cos psi 3000crss yy
psi 7325 y
So for deformation to occur the applied stress must be greater than or equal to the yield stress
Chapter 7 -
• Stronger - grain boundaries pin deformations
• Slip planes & directions (, ) change from one crystal to another.
• R will vary from one crystal to another.
• The crystal with the largest R yields first.
• Other (less favorably oriented) crystals yield later.
Adapted from Fig. 7.10, Callister 7e.(Fig. 7.10 is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].)
Slip Motion in Polycrystals
300 m
Chapter 7-
Plastic Deformation in Polycrystalline Materials
Chapter 7 -
• Can be induced by rolling a polycrystalline metal
- before rolling
235 m
- isotropic since grains are approx. spherical & randomly oriented.
- after rolling
- anisotropic since rolling affects grain orientation and shape.
rolling direction
Adapted from Fig. 7.11, Callister 7e. (Fig. 7.11 is from W.G. Moffatt, G.W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 140, John Wiley and Sons, New York, 1964.)
Anisotropy in y
Chapter 7 -
side view
1. Cylinder of Tantalum machined from a rolled plate:
rolli
ng d
irect
ion
2. Fire cylinder at a target.
• The noncircular end view shows anisotropic deformation of rolled material.
endview
3. Deformed cylinder
platethicknessdirection
Photos courtesyof G.T. Gray III,Los AlamosNational Labs. Used withpermission.
Anisotropy in Deformation
Chapter 7- 10
1. Cylinder of Tantalum machined from a rolled plate: side view
endview
• The noncircular end view shows: anisotropic deformation of rolled material.
2. Fire cylinder at a target.
3. Deformed cylinder
platethicknessdirection
Photos courtesyof G.T. Gray III,Los AlamosNational Labs. Used withpermission.
ANISOTROPY IN DEFORMATION
Chapter 7-
STRENGTHENING MECHANISMS
The ability of a metal to deform depends on the ability of dislocations to move with relative ease under loading conditions
Restricting dislocation motion will inevitably make the material stronger, need more force to induce same amount of deformation !
• Mechanisms of strengthening in single-phase metals:– grain-size reduction– solid-solution alloying– strain hardening
• Ordinarily, strengthening mechanisms reduces ductility, why ???
Chapter 7- 7
• Grain boundaries are barriers to slip.• Barrier "strength" increases with mis-orientation.
High-angle boundaries are better in blocking slip !
• Smaller grain size: more barriers to slip.
Grain size can be changed by processing !
grain boundary
slip plane
grain Agr
ain
B
yield o kyd 1/2
Adapted from Fig. 7.12, Callister 6e.(Fig. 7.12 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)
4 STRATEGIES FOR STRENGTHENING: 1: REDUCE GRAIN SIZE
Average grain size
Hall-Petch Equation :
Chapter 7- 8
• 70wt%Cu-30wt%Zn brass alloy
yield o kyd 1/2
• Data:
Adapted from Fig. 7.13, Callister 6e.(Fig. 7.13 is adapted from H. Suzuki, "The Relation Between the Structure and Mechanical Properties of Metals", Vol. II, National Physical Laboratory Symposium No. 15, 1963, p. 524.)
[grain size (mm)]-0.5
yie
ld(M
Pa)
50
100
150
200
04 8 12 16
10-1 10-2 5x10-3grain size, d (mm)
1
ky
0
GRAIN SIZE STRENGTHENING: AN EXAMPLE
Chapter 7 -
• Impurity atoms distort the lattice & generate stress.• Stress can produce a barrier to dislocation motion.
4 Strategies for Strengthening: 2: Solid Solutions
• Smaller substitutional impurity
Impurity generates local stress at A and B that opposes dislocation motion to the right.
A
B
• Larger substitutional impurity
Impurity generates local stress at C and D that opposes dislocation motion to the right.
C
D
Chapter 7-
Strategy #2: Solid Solutions
• Alloyed metals are usually stronger than their pure base metals counter parts.
Why ? Interstitial or substitutional impurities in a solution cause lattice strain, aka distortions in the lattice
Then ? • Strain field around the impurities interact with dislocation strain
fields and impede dislocation motion.• Impurities tend to diffuse and segregate around the dislocation
core to find atomic sites more suited to their radii. This reduces the overall strain energy and “anchor” the dislocation.Motion of the dislocation core away from the impurities moves it to a region of lattice where the atomic strains are greater, where lattice strains due to dislocation is no longer
compensated by the impurity atoms.
Chapter 7-
Interactions of the Stress Fields
TENSILE COMPRESSIVE
COMPRESSIVE
TENSILE
Chapter 7-
Interactions of the Stress Fields
Chapter 7 -
Stress Concentration at Dislocations
Adapted from Fig. 7.4, Callister 7e.
Chapter 7-
Impurity Segregation
Impurities tend to segregate at energetically favorable areas around the dislocation core and partially decrease the overall stress field generated around the dislocation core.
However, when stress is applied more load is needed to move dislocations with impurity atoms segregated to its core !
Chapter 7 -
Strengthening by Alloying
• small impurities tend to concentrate at dislocations• reduce mobility of dislocation increase strength
Adapted from Fig. 7.17, Callister 7e.
Chapter 7 -
Strengthening by alloying
• large impurities concentrate at dislocations on low density side
Adapted from Fig. 7.18, Callister 7e.
Chapter 7 -
Ex: Solid SolutionStrengthening in Copper
• Tensile strength & yield strength increase with wt% Ni.
• Empirical relation:
• Alloying increases y and TS.
21 /y C~
Adapted from Fig. 7.16 (a) and (b), Callister 7e.
Ten
sile
str
engt
h (M
Pa)
wt.% Ni, (Concentration C)
200
300
400
0 10 20 30 40 50 Yie
ld s
tren
gth
(MP
a)wt.%Ni, (Concentration C)
60
120
180
0 10 20 30 40 50
Chapter 7 -
• Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum).
• Result:S
~y
1
4 Strategies for Strengthening: 3: Precipitation Strengthening
Large shear stress needed to move dislocation toward precipitate and shear it.
Dislocation “advances” but precipitates act as “pinning” sites with
S.spacing
Side View
precipitate
Top View
Slipped part of slip plane
Unslipped part of slip plane
Sspacing
Chapter 7 -
• Internal wing structure on Boeing 767
• Aluminum is strengthened with precipitates formed by alloying.
Adapted from Fig. 11.26, Callister 7e. (Fig. 11.26 is courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)
1.5m
Application: Precipitation Strengthening
Adapted from chapter-opening photograph, Chapter 11, Callister 5e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)
Chapter 7 -
4 Strategies for Strengthening: 4: Cold Work (%CW)
• Room temperature deformation.• Common forming operations change the cross sectional area:
Adapted from Fig. 11.8, Callister 7e.
-Forging
Ao Ad
force
dieblank
force-Drawing
tensile force
AoAddie
die
-Extrusion
ram billet
container
containerforce
die holder
die
Ao
Adextrusion
100 x %o
do
A
AACW
-Rolling
roll
AoAd
roll
Chapter 7 -
• Ti alloy after cold working:
• Dislocations entangle with one another during cold work.• Dislocation motion becomes 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 m
Chapter 7 -
Result of Cold Work
Dislocation density =
– Carefully grown single crystal
ca. 103 mm-2
– Deforming sample increases density
109-1010 mm-2
– Heat treatment reduces density
105-106 mm-2
• Yield stress increases as d increases:
total dislocation lengthunit volume
large hardening
small hardening
y0 y1
Chapter 7- 18
• Dislocation density (d) goes up: Carefully prepared sample: d ~ 103
mm/mm3
Heavily deformed sample: d ~ 1010
mm/mm3
• Ways of measuring dislocation density:
ORlength, l1
length, l2length, l3
Volume, V
l1 l2 l3V
d dN
A
Area, A
N dislocation pits (revealed by etching)
dislocation pit
• Yield stress increases as d increases:
large hardeningsmall hardening
y0 y1
Micrograph adapted from Fig. 7.0, Callister 6e. (Fig. 7.0 is courtesy of W.G. Johnson, General Electric Co.)
40m
RESULT OF COLD WORK
Chapter 7-
• An increase in y due to plastic deformation.BUT actually # of dislocations are
increasing !!!
22
• Curve fit to the stress-strain response:
large hardening
small hardening
unlo
ad
relo
ad
y 0
y 1
T C T n“true” stress (F/A) “true” strain: ln(L/Lo)
hardening exponent: n=0.15 (some steels) to n=0.5 (some copper)
STRENGTHENING STRATEGY 4: COLD WORK (%CW)
Chapter 7-
Stress fields of dislocations interacting !The strain fields around dislocations interact with each other. Hence, they exert force on each other.
Edge dislocations, when they are in the same plane, they repel each other if they have the samesign (direction of the Burgers vector). WHY ?
They can attract and annihilate if they have opposite signs. PROVE !
Chapter 7 -
Effects of Stress at Dislocations
Adapted from Fig. 7.5, Callister 7e.
Chapter 7-
• Yield strength ( ) increases.• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.
21
y
Adapted from Fig. 7.18, Callister 6e. (Fig. 7.18 is from Metals Handbook: Properties and Selection: Iron and Steels, Vol. 1, 9th ed., B. Bardes (Ed.), American Society for Metals, 1978, p. 221.)
IMPACT OF COLD WORK
Str
ess
% cold work Strain
Chapter 7 -
Impact of Cold Work
Adapted from Fig. 7.20, Callister 7e.
• Yield strength (y) increases.• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.
As cold work is increased
Chapter 7 -
• 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
o
do
r
rrCW
Cold Work Analysis
% Cold Work
100
300
500
700
Cu
200 40 60
yield strength (MPa)
y = 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%
Chapter 7 -
• Results for polycrystalline iron:
• y and TS decrease with increasing test temperature.• %EL increases with increasing test temperature.• Why? Vacancies help dislocations move past obstacles.
Adapted from Fig. 6.14, Callister 7e.
- Behavior vs. Temperature
2. vacancies replace atoms on the disl. half plane
3. disl. glides past obstacle
-200C
-100C
25C
800
600
400
200
0
Strain
Str
ess
(M
Pa)
0 0.1 0.2 0.3 0.4 0.5
1. disl. trapped by obstacle
obstacle
Chapter 7 -
• 1 hour treatment at Tanneal... decreases TS and increases %EL.• Effects of cold work are reversed!
• 3 Annealing stages to discuss...
Adapted from Fig. 7.22, Callister 7e. (Fig.7.22 is adapted from G. Sachs and K.R. van Horn, Practical Metallurgy, Applied Metallurgy, and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, 1940, p. 139.)
Effect of Heating After %CW te
nsi
le s
tre
ngth
(M
Pa)
duc
tility
(%
EL
)tensile strength
ductility
Recovery
Recrystallization
Grain Growth
600
300
400
500
60
50
40
30
20
annealing temperature (ºC)200100 300 400 500 600 700
Chapter 7 -
Annihilation reduces dislocation density.
Recovery
• Scenario 1Results from diffusion
• Scenario 2
4. opposite dislocations
meet and annihilate
Dislocations annihilate and form a perfect atomic plane.
extra half-plane of atoms
extra half-plane of atoms
atoms diffuse to regions of tension
2. grey atoms leave by vacancy diffusion allowing disl. to “climb”
R
1. dislocation blocked; can’t move to the right
Obstacle dislocation
3. “Climbed” disl. can now move on new slip plane
Chapter 7 -
• New grains are formed that: -- have a small dislocation density -- are smalller then initial cold worked ones -- consume cold-worked grains.
Adapted from Fig. 7.21 (a),(b), Callister 7e. (Fig. 7.21 (a),(b) are courtesy of J.E. Burke, General Electric Company.)
33% coldworkedbrass
New crystalsnucleate after3 sec. at 580C.
0.6 mm 0.6 mm
Recrystallization
Chapter 7 -
• All cold-worked grains are consumed.
Adapted from Fig. 7.21 (c),(d), Callister 7e. (Fig. 7.21 (c),(d) are courtesy of J.E. Burke, General Electric Company.)
After 4seconds
After 8seconds
0.6 mm0.6 mm
Further Recrystallization
Chapter 7 -
• At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy) is reduced.
After 8 s,580ºC
After 15 min,580ºC
0.6 mm 0.6 mm
Adapted from Fig. 7.21 (d),(e), Callister 7e. (Fig. 7.21 (d),(e) are courtesy of J.E. Burke, General Electric Company.)
Grain Growth
• Empirical Relation:
Ktdd no
n elapsed time
coefficient dependenton material and T.
grain diam.at time t.
exponent typ. ~ 2
Ostwald Ripening
Chapter 7 -
TR
Adapted from Fig. 7.22, Callister 7e.
º
º
TR = recrystallization temperature
Chapter 7 -
Coldwork Calculations
A cylindrical rod of brass originally 0.40 in (10.2 mm) in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 55,000 psi (380 MPa) and a ductility of at least 15 %EL are desired. Further more, the final diameter must be 0.30 in (7.6 mm). Explain how this may be accomplished.
Chapter 7 -
Coldwork Calculations Solution
If we directly draw to the final diameter what happens?
%843100 x 400
3001100 x
4
41
100 1100 x %
2
2
2
..
.
D
D
xA
A
A
AACW
o
f
o
f
o
fo
Do = 0.40 in
BrassCold Work
Df = 0.30 in
Chapter 7 -
Coldwork Calc Solution: Cont.
• For %CW = 43.8%Adapted from Fig. 7.19, Callister 7e.
540420
y = 420 MPa– TS = 540 MPa > 380 MPa
6
– %EL = 6 < 15• This doesn’t satisfy criteria…… what can we do?
Chapter 7 -
Coldwork Calc Solution: Cont.
Adapted from Fig. 7.19, Callister 7e.
380
12
15
27
For %EL < 15
For TS > 380 MPa > 12 %CW
< 27 %CW
our working range is limited to %CW = 12-27
Chapter 7 -
Coldwork Calc Soln: RecrystallizationCold draw-anneal-cold draw again• For objective we need a cold work of %CW 12-27
– We’ll use %CW = 20• Diameter after first cold draw (before 2nd cold draw)?
– must be calculated as follows:
100
%1 100 1% 2
02
22
202
22 CW
D
Dx
D
DCW ff
50
02
2
100
%1
.
f CW
D
D
50
202
100%
1.
f
CW
DD
m 335010020
130050
021 ..DD.
f
Intermediate diameter =
Chapter 7 -
Coldwork Calculations Solution
Summary:
1. Cold work D01= 0.40 in Df1 = 0.335 m
2. Anneal above D02 = Df1
3. Cold work D02= 0.335 in Df 2 =0.30 m
Therefore, meets all requirements
20100 3350
301%
2
2
x
.
.CW
24%
MPa 400
MPa 340
EL
TSy
%CW1 10.335
0.4
2
x 100 30
Fig 7.19
Chapter 7 -
Rate of Recrystallization
• Hot work above TR
• Cold work below TR
• Smaller grains – stronger at low temperature– weaker at high temperature
t/RT
BCt
kT
ERtR
1:note
log
logloglog 0
RT1
log t
start
finish
50%
TR ~ 0.3 Tm-0.7 Tm
TR depends on %CW•Decrease with increasing % CW
No Strain hardening occurs
Chapter 7 -
• Dislocations are observed primarily in metals and alloys.
• Strength is increased by making dislocation motion difficult.
• Particular ways to increase strength are to: --decrease grain size --solid solution strengthening --precipitate strengthening --cold work
• Heating (annealing) can reduce dislocation density and increase grain size. This decreases the strength.
Summary
Chapter 7-
Reading: Chapter 7
Core Problems: 7.6, 7.7, 7.13 (see page 183-4), 7.14, 7.23, 7.31, 7.41 (figure 7.25)
Bonus Problems: 7.40, 7.36, 7.39
Due date is April 3th
0
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