1. Chapter 7: Dislocations & 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?

2. Dislocations & Materials Classes
Dislocation types: edge and screw
• Metals: Disl. motion easier.
-non-directional bonding
-close-packed directions
for slip.
electron cloud
ion cores +
• Covalent Ceramics
(Si, diamond): Motion hard.
- very hard, Tm>3550C
-directional bonding
• Ionic Ceramics (NaCl):
Motion hard.
-need to avoid ++ and - -
neighbors.
- non-directional bonding + -
PD corresponds to the motion of large numbers of dislocations

3. Dislocation Motion
Plastic deformation mechanisms are 1. slipping 2. twinning
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).
So we saw that above the yield stress plastic deformation occurs. But how? In a perfect single crystal for this to occur every bond connecting tow planes would have to break at once! Large energy requirement
Now rather than entire plane of bonds needing to be broken at once, only the bonds along dislocation line are broken at once.
If dislocations don't move, deformation doesn't occur!

4. Dislocation Motion
Slip is the process by which plastic deformation is produced by dislocation motion
Dislocation moves along slip plane in slip direction perpendicular to dislocation line( slip system)
Slip direction have the same direction as Burgers vector
Edge dislocation
Screw dislocation

5. Slipping Mechanism Slip System
Slip plane – plane along which the dis. line traverses
plane allowing easiest slippage
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

6. The slip planes are usually the most densely packed planes
The slip planes are usually the most densely packed planes. Slip is favored on closed packed planes since a lower shear stress for atomic displacement is required than for less densely packed planes.
Slip in the closed packed direction is also favored since less energy is required to move the atoms from one position to another if the atoms are closer together.

7. Slip: The process by which plastic deformation is produced by dislocation motion
The slip system depends on the crystal structure of the metal and is such that the atomic distortion that accompanies the motion of a dislocation is a minimum
dislocation density: (The number of dislocations). is expressed as the total dislocation length per unit volume, or equivalently, the number of dislocations that intersect a unit area of a random section.
The units of dislocation density are millimeters of dislocation per cubic
millimeter or just per square millimeter. Dislocation densities as low as 10 3 mm -2
All metals and alloys contain some dislocations that were introduced during 1.solidification, 2.during plastic deformation, and as a consequence of 3.thermal stresses that result from rapid cooling.

8. Stress and Dislocation Motion
• Crystals slip due to a resolved shear stress, tR.
• Applied tension can produce such a stress.
Applied tensile
stress:
= F/A s
direction
slip F A
slip plane
normal, ns
Resolved shear
stress: tR = F s /A
direction
slip AS FS
direction
slip
Relation between s
and tR = FS
/AS F
cos l A / f nS AS

9. Critical Resolved Shear Stress
• Condition for dislocation motion:
Critical resolved shear stress: is the minimum shear stress required to initiate slip and is a property of the material that determine when yielding occurs
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

10. For
The crystal ordinarily fractures rather than deforming plastically

11. Single Crystal Slip

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

13. Ex: Deformation of single crystal
What stress is necessary (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

14. Slip Motion in Polycrystals
300 mm
• Stronger metals- grain boundaries pin deformations
• 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.

15. Anisotropy in sy
Isotropic. Having identical values of a property in all crystallographic directions.
• Can be induced by rolling a polycrystalline metal
- before rolling
- after rolling
- anisotropic
since rolling affects grain
orientation and shape.
rolling direction
235 mm
- isotropic
since grains are
approx. spherical
& randomly
oriented.

16. Anisotropy in Deformation
• The noncircular end view shows
anisotropic deformation of rolled material.
end
view
3. Deformed
cylinder
plate
thickness
direction
1. Cylinder of
Tantalum
machined
from a
rolled plate:
rolling direction
2. Fire cylinder
at a target.
side view

17. Twinning mechanism
A part of the atomic lattice is deformed so that is forms a mirror image of the un-deformed lattice next to it.
Twinning plane: is the plane between the un-deformed and deformed parts of the metal lattice

18. Comparison between slip and twinning mechanisms
slipping
twinning
the atoms in one side of the slip plane all move equal distances
the atoms move distances proportional to their distance from the twinning plane
Slip Leaves a series of
steps (lines)
2. Twinning leaves small but well defined regions of the crystal deformed
3. Most for FCC and BCC structure, they have more slip systems
3. Is most important for HCP structure , because its small number of slip system
4. normally slip results in relatively large deformations
4. only small deformations result for twinning

19. Mechanisms of strengthening in metals
The ability of a metal to plastically deform depends on the ability of dislocations to move.
ductility is sacrificed when an alloy is strengthened
restricting or hindering dislocation motion renders a material harder and stronger.
strengthening mechanisms for single phase metals, by grain size reduction, solid-solution alloying, and strain hardening(cold work).

20. 4 Strategies for Strengthening: 1: Reduce Grain Size
• Grain boundaries are
barriers to slip.
Change direction
Discontinuity of slip
planes
• Barrier "strength"
increases with
Increasing angle of
mis-orientation.
• Smaller grain size:
more barriers to slip.
• Hall-Petch Equation:
d = average grain diameter
σy - yield strength
σ0, ky are constant for particular material

21. Materials are a fine grained materials (one that has small grains) and a coarse grained. Fine grained material is harder and stronger than coarse grained.
Grain size may be regulated by
the rate of solidification from the liquid phase, and
also by plastic deformation followed by an appropriate heat treatment ( depends on time and temperature). The greater tem. And time the greater grain size.

22. 4 Strategies for Strengthening: 2: Solid Solutions
• Impurity atoms distort the lattice & generate stress.
• Stress can produce a barrier to dislocation motion.
• 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

23. Stress Concentration at Dislocations
Don’t move past one another – hardens material

24. Strengthening by Alloying
small impurities tend to concentrate at dislocations
reduce mobility of dislocation  increase strength

25. Strengthening by alloying
large impurities concentrate at dislocations on low density side

26. 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
• Empirical relation:
• Alloying increases sy and TS.

27. 4 Strategies for Strengthening: 3: Cold Work (%CW)
Strain hardening: a ductile metal becomes harder and stronger as it is plastically deformed( the effect of CW may be removed by annealing HT)
• Room temperature deformation.
• Common forming operations change the cross
sectional area:
-Forging A o d
force
die
blank
-Rolling
roll A o d
-Drawing
tensile
force A o d
die
-Extrusion
ram
billet
container
force
die holder
die A o d
extrusion

28. 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.

29. 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
Again it propagates through til reaches the edge
large hardening
small hardening s e y0 y1
• Yield stress increases
as rd increases:

30. Effects of Stress at Dislocations

31. Impact of Cold Work As cold work is increased
• Yield strength (sy) increases.
• Tensile strength (TS) increases.
• Ductility (%EL or %AR) decreases.

32. Cold Work Analysis Cu Cu Cu • What is the tensile strength &
=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%

33. 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
• 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

34. 4 Strategies for Strengthening: 4: Precipitation Strengthening
• Hard precipitates are difficult to shear.
Ex: Ceramics in metals (SiC in Iron or Aluminum).
Side View
precipitate
Top View
Slipped part of slip plane
Unslipped part of slip plane S
spacing
Large shear stress needed
to move dislocation toward
precipitate and shear it.
Dislocation
“advances” but
precipitates act as
“pinning” sites with
spacing S .
• Result:

35. Application: Precipitation Strengthening
• Internal wing structure on Boeing 767
• Aluminum is strengthened with precipitates formed
by alloying.
1.5mm

36. Note: The strengthening effects due to grain size reduction and strain hardening can be eliminated or at least reduced by an elevated temperature heat treatment. Conversely, solid solution strengthening is unaffected by heat treatment

37. Recovery, Recrystallization, and Grain Growth
These properties and structures may revert back to the precold-worked states
by appropriate heat treatment (sometimes termed an annealing treatment). Such
restoration results from two different processes that occur at elevated temperatures:
recovery and recrystallization, which may be followed by grain growth.

38. Effect of Heating After %CW
• 1 hour treatment at Tanneal...
decreases TS and increases %EL.
• Effects of cold work are reversed!
tensile strength (MPa)
ductility (%EL)
tensile strength
ductility
Recovery
Recrystallization
Grain Growth
600
300
400
500 60 50 40 30 20
annealing temperature (ºC)
200
100
700
• 3 Annealing
stages to
discuss...

39. 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

40. 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

41. Further Recrystallization
• All cold-worked grains are consumed.
After 4
seconds
After 8
0.6 mm

42. TR = recrystallization temperatureTR
TR = recrystallization temperature

43. 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

44. Grain Growth
grain growth: After recrystallization is complete, the strain-free grains will continue to grow if the metal specimen is left at the elevated temperature this phenomenon is called grain growth.
Grain growth does not need to be preceded by recovery and recrystallization; it may
occur in all polycrystalline materials, metals and ceramics alike
As grains increase in size, the total boundary area decreases, yielding an attendant
reduction in the total energy; this is the driving force for grain growth.
Grain growth occurs by the migration of grain boundaries. Obviously, not all
grains can enlarge, but large ones grow at the expense of small ones that shrink.
The directions of boundary movement and atomic motion are opposite to each
other
growth via atomic diffusion

45. Grain Growth • At longer times, larger grains consume smaller ones.
• Why? Grain boundary area (and therefore energy)
is reduced.
After 8 s,
580ºC
After 15 min,
0.6 mm
• Empirical Relation:
elapsed time
coefficient dependent
on material and T.
grain diam.
at time t.
exponent typ. ~ 2

46. 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.

47. Coldwork Calculations Solution
If we directly draw to the final diameter what happens?
Brass
Cold
Work D f
= 0.30 in D o
= 0.40 in

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

49. Coldwork Calc Solution: Cont.
380 12 15 27
For TS > 380 MPa
> 12 %CW
For %EL < 15
< 27 %CW
 our working range is limited to %CW = 12-27

50. Coldwork Calc Soln: Recrystallization
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: 
So after the cold draw & anneal D02=0.335m
Intermediate diameter =

51. Coldwork Calculations Solution
Summary:
Cold work D01= 0.40 in  Df1 = m
Anneal above D02 = Df1
Cold work D02= in  Df 2 =0.30 m
Therefore, meets all requirements
Fig 7.19 

52. Rate of Recrystallization
Hot work  above TR
Cold work  below TR
Smaller grains
stronger at low temperature
weaker at high temperature
log t
start
finish
50%

53. 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.

54. problem
The lower yield point for an iron that has an average grain diameter of 1x10-2 mm is 230 MPa (33,000 psi). At a grain diameter of 6x10-3 mm, the yield point increases to 275 MPa (40,000 psi). At what grain diameter will the lower yield point be 310 MPa (45,000 psi)?

55. We are asked to determine the grain diameter for an iron which will give a yield strength of 310 MPa (45,000 psi). The best way to solve this problem is to first establish two simultaneous expressions of Equation 7.7, solve for σ0 and ky, and finally determine the value of d when σy = 310 MPa. The data pertaining to this problem may be tabulated as follows:

56. problem
The critical resolved shear stress for copper is 0.48 MPa (70 psi). Determine the maximum possible yield strength for a single crystal of Cu pulled in tension.
Answer:
In order to determine the maximum possible yield strength for a single crystal of Cu pulled in tension, we simply employ Equation 7.5 as

57. problem
Two previously un-deformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 15 mm and 12 mm, respectively. The second specimen, with an initial radius of 11 mm, must have the same deformed hardness as the first specimen; compute the second specimen’s radius after deformation.

58. In order for these two cylindrical specimens to have the same deformed hardness, they must be deformed to the same percent cold work. For the first specimen
For the second specimen, the deformed radius is computed using the above equation and solving for rd as