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CHAPTER 8: DEFORMATION AND STRENGTHENING MECHANISMS. ISSUES TO ADDRESS. • Why are dislocations observed primarily in metals and alloys?. • How are strength and dislocation motion related?. • How do we increase strength?. • How can heating change strength and other properties?. 1. - PowerPoint PPT Presentation
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ISSUES TO ADDRESS...• Why are dislocations observed primarily in metals and alloys?
• How are strength and dislocation motion related?
• How do we increase strength?
1
• How can heating change strength and other properties?
CHAPTER 8: DEFORMATION AND
STRENGTHENING MECHANISMS
3
• Produces plastic deformation,• Depends on incrementally breaking bonds.
Plasticallystretchedzincsinglecrystal.
• If dislocations don't move, deformation doesn't happen!
DISLOCATION MOTION
+ +
+ +
+ + + + + + + + + + + + +
+ + + + + + +
2
• Metals: Disl. motion easier. -non-directional bonding -close-packed directions for slip.
electron cloud ion cores
• Covalent Ceramics (Si, diamond): Motion hard. -directional (angular) bonding• Ionic Ceramics (NaCl): Motion hard. -need to avoid ++ and -- neighbors.
DISLOCATIONS & MATERIALS CLASSES
+ + + +
+ + +
+ + + +
- - -
- - - -
- - -
6
• Structure: close-packed planes & directions are preferred.
• Comparison among crystal structures: FCC: many close-packed planes/directions; HCP: only one plane, 3 directions; BCC: none
close-packed plane (bottom)close-packed plane (top)
close-packed directions
Mg (HCP)
Al (FCC)tensile direction
• Results of tensile testing.
view onto twoclose-packedplanes.
DISLOCATIONS & CRYSTAL STRUCTURE
7
• Crystals slip due to a resolved shear stress, R. • Applied tension can produce such a stress.
Rcoscos
Relation between and R
R=Fs/As
Fcos A/cosns
AAs
STRESS AND DISLOCATION MOTION
Applied tensile stress: = F/A
FA
Fsli
p
direct
ion
Resolved shear stress: R=Fs/As
As
R
R
Fs
slip
direct
ion
slip plane
normal, ns
F
Fssli
p
direct
ion
8
• Condition for dislocation motion: R CRSS
• Crystal orientation can make it easy or hard to move disl.
10-4G to 10-2G
typically
Rcoscos
CRITICAL RESOLVED SHEAR STRESS
R = 0
=90°
R = /2=45°=45°
R = 0
=90°
9
• 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.
300 m
DISL. MOTION IN POLYCRYSTALS
10
• Grain boundaries are barriers to slip.• Barrier "strength" increases with misorientation.• Smaller grain size: more barriers to slip.
• Hall-Petch Equation:
grain boundary
slip plane
grain Agr
ain
B
yield o kyd 1/2
4 STRATEGIES FOR STRENGTHENING: 1: REDUCE
GRAIN SIZE
11
• 70wt%Cu-30wt%Zn brass alloy
yield o kyd 1/2
• Data:
[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
0.75mm
GRAIN SIZE STRENGTHENING: AN EXAMPLE
• Can be induced by rolling a polycrystalline metal
12
-before rolling -after rolling
235 m
-isotropic since grains are approx. spherical & randomly oriented.
-anisotropic since rolling affects grain orientation and shape.
rolling direction
ANISOTROPY IN yield
14
• Impurity atoms distort the lattice & generate stress.• Stress can produce a barrier to dislocation motion.• Smaller substitutional impurity
• Larger substitutional impurity
Impurity generates local shear at A and B that opposes disl motion to the right.
Impurity generates local shear at C and D that opposes disl motion to the right.
STRENGTHENING STRATEGY 2: SOLID SOLUTIONS
C
D
A
B
15
• Tensile strength & yield strength increase w/wt% Ni.
• Empirical relation:
• Alloying increases y and TS. y ~C1/2
Yield
str
ength
(M
Pa)
wt. %Ni, (Concentration C)
60
120
180
0 10 20 30 40 50
Tensi
le s
trength
(M
Pa)
wt. %Ni, (Concentration C)
200
300
400
0 10 20 30 40 50
EX: SOLID SOLUTIONSTRENGTHENING IN COPPER
16
• Room temperature deformation.• Common forming operations change the cross sectional area:
%CW
Ao AdAo
x100
Ao Ad
force
dieblank
force
-Forging -Rolling
-Extrusion-Drawing
tensile force
AoAddie
dieram billet
container
containerforce
die holder
die
Ao
Adextrusion
roll
AoAd
roll
STRENGTHENING STRATEGY : COLD WORK (%CW)
17
• Ti alloy after cold working:
• Dislocations entangle with one another during cold work.• Dislocation motion becomes more difficult.
0.9 m
DISLOCATIONS DURING COLD WORK
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
dd
NA
Area, A
N dislocation pits (revealed by etching)
dislocation pit
• Yield stress increases as d increases:
large hardeningsmall hardening
y0 y1
40m
RESULT OF COLD WORK
• Dislocation generate stress.• This traps other dislocations.
20
DISLOCATION-DISLOCATION TRAPPING
Red dislocation generates shear at pts A and B that opposes motion of green disl. from left to right.
A
B
Str
ess
% cold work Strain
• Yield strength ( ) increases.• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.
21
y
IMPACT OF COLD WORK
• What is the tensile strength & ductility after cold working?
22
Cold work ----->
Do=15.2mm Dd=12.2mm
Copper
%CW ro
2 rd2
ro2
x10035.6%
COLD WORK ANALYSIS
ductility (%EL)
7%
%EL=7%% Cold Work
20
40
60
20 40 6000
Cu
% Cold Worky=300MPa
100
300
500
700
Cu
200 40 60
yield strength (MPa)
300MPa
% Cold Work
tensile strength (MPa)
340MPa
TS=340MPa
200Cu
0
400
600
800
20 40 60
• Results for polycrystalline iron:
23
• y and TS decrease with increasing test temperature.• %EL increases with increasing test temperature.• Why? Vacancies help dislocations past obstacles.
1. disl. trapped by obstacle
2. vacancies replace atoms on the disl. half plane
3. disl. glides past obstacle
obstacle
- BEHAVIOR VS TEMPERTURE
00 0.1 0.2 0.3 0.4 0.5
200
400
600
800
Str
ess
(M
Pa)
Strain
-200°C
-100°C
25°C
• 1 hour treatment at Tanneal... decreases TS and increases %EL.• Effects of cold work are reversed!
24
• 3 Annealing stages to discuss...
EFFECT OF HEATING AFTER %CW
ten
sile
str
en
gth
(M
Pa)
du
ctilit
y (
%E
L)
Annealing Temperature (°C)
300
400
500
600 60
50
40
30
20Recovery
Recrystallization
Grain Growth
ductility
tensile strength300 700500100
Annihilation reduces dislocation density.
25
• Scenario 1
• Scenario 2
atoms diffuse to regions of tension
extra half-plane of atoms
extra half-plane of atoms
Disl. annhilate and form a perfect atomic plane.
1. dislocation blocked; can’t move to the right
obstacle dislocation
2. grey atoms leave by vacancy diffusion allowing disl. to “climb”
4. opposite dislocations meet and annihilate
3. “Climbed” disl. can now move on new slip plane
R
RECOVERY
• New crystals are formed that: --have a small disl. density --are small --consume cold-worked crystals.
26
33% coldworkedbrass
New crystalsnucleate after3 sec. at 580C.
0.6 mm 0.6 mm
RECRYSTALLIZATION
• All cold-worked crystals are consumed.
27
After 4seconds
After 8seconds
0.6 mm0.6 mm
FURTHER RECRYSTALLIZATION
• At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy) is reduced.
28
• Empirical Relation:
After 8 s,580C
After 15 min,580C
dn do
n Ktelapsed time
coefficient dependenton material and T.
grain diam.at time t.
exponent typ. ~ 2
0.6 mm 0.6 mm
GRAIN GROWTH
Example
Using the diagram, compute the time for the average grain size to increase form 0.01-mm to 0.1-mm at (a) 500 oC and (b) 600 oC
29
0
unload/reload
0
brittle failure
plastic failure
20
40
60
2 4 6
(MPa)
x
x
semi- crystalline
case
amorphous regions elongate
crystalline regions align
crystalline regions
slide
8
onset of necking
aligned, cross- linked case
networked case
Initial
Near Failure
near failure
TENSILE RESPONSE: BRITTLE & PLASTIC
30
• Drawing... --stretches the polymer prior to use --aligns chains to the stretching direction• Results of drawing: --increases the elastic modulus (E) in the stretching dir. --increases the tensile strength (TS) in the stretching dir. --decreases ductility (%EL)• Annealing after drawing... --decreases alignment --reverses effects of drawing.• Compare to cold working in metals!
Adapted from Fig. 15.12, Callister 6e. (Fig. 15.12 is from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp. 500-501.)
PREDEFORMATION BY DRAWING
31
• Thermoplastics: --little cross linking --ductile --soften w/heating --polyethylene (#2) polypropylene (#5) polycarbonate polystyrene (#6)
• Thermosets: --large cross linking (10 to 50% of mers) --hard and brittle --do NOT soften w/heating --vulcanized rubber, epoxies, polyester resin, phenolic resin
Callister, Fig. 16.9
T
Molecular weight
Tg
Tmmobile liquid
viscous liquid
rubber
tough plastic
partially crystalline solid
crystalline solid
THERMOPLASTICS VS THERMOSETS
32
• Compare to responses of other polymers: --brittle response (aligned, cross linked & networked case) --plastic response (semi-crystalline case)
TENSILE RESPONSE: ELASTOMER CASE
initial: amorphous chains are kinked, heavily cross-linked.
final: chains are straight,
still cross-linked
0
20
40
60
0 2 4 6
(MPa)
8
x
x
x
elastomer
plastic failure
brittle failure
Deformation is reversible!
33
• Dislocations are observed primarily in metals and alloys.• Here, 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.
SUMMARY