1. Introduction to Materials Science and Engineering
Chapter 9 (Textbook):
Dislocations and Strengthening Mechanisms

2. Content Introduction Dislocations and Plastic Deformation
Strengthening in Metals
Recrystallization and Grain Growth

3. ISSUES TO ADDRESS...
• Why are dislocations observed primarily in metals and alloys?
• How are the strength and dislocation motion related?
• How do we increase strength?
• How can heating change strength and other properties?

4. Content Introduction Dislocations and Plastic Deformation
Strengthening in Metals
Recrystallization and Grain Growth

5. Dislocation Density Carefully solidified metal 103 mm-2
참고자료
Dislocation Density
Carefully solidified metal 103 mm-2
Heavily deformed metals 109 ~ 1010 mm-2
Heavily deformed metals after heat treatment 105 ~ 106 mm-2
Ceramics 102 ~ 104 mm-2
Si wafer 0.1 ~ 1 mm-2

6. Dislocations & Materials Classes
Metals: Dislocation motion is easier.
- non-directional bonding
- close-packed planes and 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. -

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.

8. Dislocation Motion
 Dislocation moves along slip plane in slip direction perpendicular to dislocation line
 Slip deformation is the same direction as Burgers vector
Edge dislocation
Adapted from Fig. 7.2, Callister 7e.
Screw dislocation

9. Stress field of edge dislocation

10. Forces between dislocations
참고자료
Two like edge dislocations (i.e., both have parallel Burgers vectors) lying on the same slip plane will repel each other.
So why? – think about energetics

11. Forces between dislocations
참고자료
Two unlike edge dislocations (i.e., both have opposite Burgers vectors) lying on the same slip plane will attract each other and annihilate out.
So why? – thick about energetics

12. Effects of Stress at Dislocations
Adapted from Fig. 7.5, Callister 7e.

13. Deformation Mechanisms
Slip System:
- Slip plane - plane allowing easiest slippage
Wide interplanar spacing - highest planar density
- Slip direction - direction of movement - highest linear density
- 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 (refer class-note in chapter 3)
Adapted from Fig. 7.6, Callister 7e.

14. Stress and Dislocation Motion
Crystals slip due to a resolved shear stress, tR.
Applied tension can produce such a stress.

15. Stress and Dislocation Motion
Applied tensile stress
direction
slip F A
slip plane
normal, ns
Resolved shear stress
direction
slip AS tR FS
direction
slip
Relation between s and tR l F FS f nS AS A
where,

16. Critical Resolved Shear Stress
Condition for dislocation motion:
10-4 GPa to 10-2 GPa
typically
Crystal orientation can make
it easy or hard to move dislocation tR
= 0 l
=90° s tR = s /2 l
=45° f tR
= 0 f
=90° s
 maximum at  =  = 45º

17. Single Crystal Slip Adapted from Fig. 7.9, Callister 7e.

18. Ex: Deformation of single crystal
a) Will the single crystal yield?
b) If not, what stress is needed?
=60°
crss = 3000 psi
=35°
Adapted from Fig. 7.7, Callister 7e.
 = 6500 psi
So the applied stress of 6500 psi will not cause the crystal to yield.

19. Ex: Deformation of single crystal
Then, what stress is necessary for dislocation to move (i.e., what is the yield stress, sy)?
So for deformation to occur the applied stress must be greater than or equal to the yield stress

20. Slip Motion in Polycrystals
300 mm
Slip planes & directions
(l, f) change from one
crystal to another.
tR will vary from one
The crystal with the
largest tR yields first.
Other (less favorably
oriented) crystals
yield later.
Adapted from Fig. 7.10, Callister 7e.
(Fig is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].)

21. Anisotropy in sy • Can be induced by rolling a polycrystalline metal
- before rolling
- after rolling
- anisotropic
since rolling affects grain
orientation and shape.
rolling direction
Adapted from Fig. 7.11, Callister 7e. (Fig 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.)
235 mm
- isotropic
since grains are
approx.
spherical
& randomly oriented.

22. Content Introduction Dislocations and Plastic Deformation
Strengthening in Metals
Recrystallization and Grain Growth

23. Strategies for Strengthening:
Any method for dislocation not to move:
Grain Size Reduction
Solid Solution Strengthening
Precipitation Strengthening
Cold Work (strain hardening)

24. Strategies for Strengthening: 1: Reducing the Grain Size
Grain boundaries are barriers
to slip.
Barrier "strength"
increases with
Increasing the angle of
misorientation.
Smaller grain size:
more barriers to slip.
Hall-Petch Equation:
Adapted from Fig. 7.14, Callister 7e.
(Fig is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

25. Grain Size Reduction
70wt%Cu-30wt%Zn brass alloy
0.75mm
Data:

26. Strategies for Strengthening: 2: Solid Solutions
Impurity atoms distort the lattice & generate stress.
Stress can produce a barrier to dislocation motion.
small impurities tend to concentrate at dislocations
reduce mobility of dislocation  increase strength
 Tensile lattice strain
Adapted from Fig. 7.17, Callister 7e.

27. Strengthening by alloying
large impurities concentrate at dislocations on low density side
 Compressive lattice strain

28. Ex: Solid Solution Strengthening in Copper
Tensile strength & yield strength increase with wt% Ni.
Tensile strength (MPa)
wt.% Ni, (Concentration C)
200
300
400 10 20 30 40 50
Yield strength (MPa)
wt.%Ni, (Concentration C) 60
120
180 10 20 30 40 50
Adapted from Fig (a) and (b), Callister 7e.
Empirical relation:
Alloying increases sy and TS.

29. Strategies for Strengthening: 3: Precipitation Strengthening
Hard precipitates are difficult to shear.
Ex: Ceramics in metals (SiC in Iron or Aluminum).
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
Result:

30. Simulation: Precipitation Strengthening
참고자료
Simulation: Precipitation Strengthening
View onto slip plane of Nimonic PE16
Precipitate volume fraction: 10%
Average precipitate size: 64 b (b = 1 atomic slip distance)

31. Application: Precipitation Strengthening
참고자료
Application: Precipitation Strengthening
Internal wing structure on Boeing 767
Adapted from chapter-opening photograph, Chapter 11, Callister 5e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)
Aluminum is strengthened with precipitates formed
by alloying.
1.5mm
Adapted from Fig , Callister 7e.
(Fig is courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

32. 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 A o d
force
die
blank
-Drawing
tensile
-Extrusion
ram
billet
container
die holder
extrusion
-Rolling
roll

33. Dislocations During Cold Work
Ti alloy after cold working:
0.9 mm
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.)

34. Result of Cold Work total dislocation length Dislocation density =
unit volume
Dislocation density =
- Carefully grown single crystal
 ca. 103 mm-2
- Deforming sample increases density
 mm-2
- Heat treatment reduces density
 mm-2
Yield stress increases
as rd increases:
large hardening
small hardening s e y0 y1

35. Impact of Cold Work As cold work is increased
Yield strength (sy) increases.
Tensile strength (TS) increases.
Ductility (%EL or %AR) decreases.
As cold work is increased
Adapted from Fig. 7.20, Callister 7e.

36. Cold Work Analysis Cu Cu Cu What is the tensile strength &ductility
=12.2mm
Copper
What is the tensile strength &ductility
after cold working? D o
=15.2mm
% Cold Work
100
300
500
700 Cu 20 40 60
yield strength (MPa)
% Cold Work
tensile strength (MPa)
200 Cu
400
600
800 20 40 60
ductility (%EL)
% Cold Work 20 40 60 Cu s y
= 300MPa
300MPa
340MPa
TS = 340MPa 7%
%EL
= 7%
Adapted from Fig. 7.19, Callister 7e. (Fig 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.)

37. Content Introduction Dislocations and Plastic Deformation
Strengthening in Metals
Recrystallization and Grain Growth

38. s- e Behavior vs. Temperature
Results for
polycrystalline iron:
-200C
-100C
25C
800
600
400
200
Strain
Stress (MPa)
0.1
0.2
0.3
0.4
0.5
Adapted from Fig. 6.14, Callister 7e.
sy and TS decrease with increasing test temperature.
%EL increases with increasing test temperature.
Why? Vacancies
help dislocations
move past obstacles. 3
. disl. glides past obstacle
2. vacancies
replace
atoms on the
disl. half
plane
1. disl. trapped
by obstacle
obstacle

39. Effect of Heating after CW
1 hour treatment at Tanneal decreases TS and increases %EL
Effects of cold work are reversed
ductility
Tensile strength
three annealing stages

40. Recovery tR Annihilation reduces dislocation density. Scenario 1
Results from diffusion
extra half-plane
of atoms
Dislocations
annihilate
and form
a perfect
atomic
plane.
atoms
diffuse
to regions
of tension
Scenario 2 tR
1. dislocation blocked;
can’t move to the right
Obstacle dislocation 3
. “Climbed” disl. can now
move on new slip plane 2
. grey atoms leave by
vacancy diffusion
allowing disl. to “climb”
4. opposite dislocations
meet and annihilate

41. Recrystallization New grains are formed that:
- have a small dislocation density
- are small
- consume cold-worked grains.
33% cold
worked
brass
New crystals
nucleate after
3 sec. at 580C.
0.6 mm
Adapted from Fig (a),(b), Callister 7e. (Fig (a),(b) are courtesy of J.E. Burke, General Electric Company.)

42. Further Recrystallization
All cold-worked grains are consumed.
After 4
seconds
After 8
0.6 mm
Adapted from Fig (c),(d), Callister 7e. (Fig (c),(d) are courtesy of J.E. Burke, General Electric Company.)

43. Grain Growth At longer times, larger grains consume smaller ones.
Why? Grain boundary area (= energy) is reduced.
After 8 s,
580ºC
After 15 min,
0.6 mm
Adapted from Fig (d),(e), Callister 7e. (Fig (d),(e) are courtesy of J.E. Burke, General Electric Company.)
Empirical Relation:
elapsed time
coefficient dependent
on material and T.
grain diameter
at time t.
exponent typ. ~ 2
Ostwald Ripening

44. Process: TR = recrystallization temperature TR º º
Adapted from Fig. 7.22, Callister 7e.

45. Recrystallization Temperature, TR
TR = recrystallization temperature = point of highest rate of property change
Tm => TR  Tm (K)
Due to diffusion  annealing time TR = f(t) shorter annealing time => higher TR
Higher %CW => lower TR – strain hardening
Pure metals lower TR due to dislocation movements
- Easier to move in pure metals => lower TR

46. Cold work 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. Furthermore, the final diameter must be 0.30 in (7.6 mm). Explain how this may be accomplished.
Step 1: If we directly draw to the final diameter what happens? D o
= 0.40 in
Brass
Cold
Work f
= 0.30 in

47. Coldwork Calculations: Cont.
420
540 6
For %CW = 43.8%
Adapted from Fig. 7.19, Callister 7e.
- y = 420 MPa
- TS = 540 MPa > 380 MPa
- %EL = < 15
This doesn’t satisfy criteria…… what can we do?

48. Coldwork Calculations: Cont.
380 12 15 27
Adapted from Fig. 7.19, Callister 7e.
For TS > 380 MPa
> 12 %CW
For %EL < 15
< 27 %CW
 our working range is limited to %CW = 12-27

49. Intermediate diameter:
Cold work Calculations:
Cold 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: 
Intermediate diameter:

50. Cold work Calculations:
Summary:
Cold work D0= 0.40 in  Df1 = in
Anneal above Df1
Cold work Df1= in  Df 2 =0.30 in
Therefore, meets all requirements
Fig 7.19 

51. Rate of Recrystallization
Hot work  above TR
Cold work  below TR
log t
start
finish
50%

52. Summary 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.
- recovery
- recrystallization
- grain growth