1. Dislocations & Strengthening Mechanisms
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Chapter 7:
Dislocations & Strengthening Mechanisms
Dr. Mohammad Abuhaiba, PE

2. WHY STUDY Dislocations and Strengthening Mechanisms?
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WHY STUDY Dislocations and Strengthening Mechanisms?
Understand underlying mechanisms of the techniques used to strengthen and harden metals and their alloys.
Design and tailor mechanical properties of materials.
Dr. Mohammad Abuhaiba, PE

3. WHY STUDY Dislocations and Strengthening Mechanisms?
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WHY STUDY Dislocations and Strengthening Mechanisms?
Strengthening mechanisms will be applied to the development of mechanical properties for steel alloys.
Why heat treating a deformed metal alloy makes it softer and more ductile and produces changes in its microstructure.
Dr. Mohammad Abuhaiba, PE

4. Learning Objectives Describe edge and screw dislocation motion.
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Learning Objectives
Describe edge and screw dislocation motion.
Describe how plastic deformation occurs by motion of edge & screw dislocations in response to applied shear stresses.
Define slip system.
Describe how grain structure of a polycrystalline metal is altered when it is plastically deformed.
Explain how grain boundaries impede dislocation motion and why a metal having small grains is stronger than one having large grains.
Dr. Mohammad Abuhaiba, PE

5. 4/22/2017 3:35 AM
Learning Objectives
Describe & explain solid-solution strengthening for substitutional impurity atoms in terms of lattice strain interactions with dislocations.
Describe and explain phenomenon of strain hardening in terms of dislocations and strain field interactions.
Describe recrystallization in terms of both alteration of microstructure and mechanical characteristics of the material.
Describe phenomenon of grain growth from both macroscopic and atomic perspectives.
Dr. Mohammad Abuhaiba, PE

6. 7.2 BASIC CONCEPTS Dislocation Motion
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7.2 BASIC CONCEPTS Dislocation Motion
Cubic & hexagonal metals - plastic deformation is by plastic shear or slip where one plane of atoms slides over adjacent plane by dislocations motion.
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.
Dr. Mohammad Abuhaiba, PE

7. 7.3 Characteristics of Dislocations
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7.3 Characteristics of Dislocations
When metals are plastically deformed:
Around 5% of deformation energy is retained internally
remainder is dissipated as heat
The major portion of stored energy is strain energy associated with dislocations.
Dr. Mohammad Abuhaiba, PE

8. 7.3 Characteristics of Dislocations
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7.3 Characteristics of Dislocations
Figure 7.4: some atomic lattice distortion exists around the dislocation line because of presence of extra half-plane of atoms.
Don’t move past one another – hardens material
Dr. Mohammad Abuhaiba, PE

9. 7.3 Characteristics of Dislocations
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7.3 Characteristics of Dislocations
Atoms immediately above and adjacent to dislocation line are squeezed together. As a result, these atoms may be thought of as experiencing a compressive strain relative to atoms positioned in the perfect crystal and far removed from the dislocation.
Directly below the half-plane, the lattice atoms sustain an imposed tensile strain.
Dr. Mohammad Abuhaiba, PE

10. 7.3 Characteristics of Dislocations
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7.3 Characteristics of Dislocations
Shear strains also exist in the vicinity of the edge dislocation.
For a screw dislocation, lattice strains are pure shear only.
These lattice distortions may be considered to be strain fields that radiate from the dislocation line.
The strains extend into the surrounding atoms, and their magnitude decreases with radial distance from the dislocation.
Dr. Mohammad Abuhaiba, PE

11. 7.3 Characteristics of Dislocations strain fields surrounding
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7.3 Characteristics of Dislocations strain fields surrounding
Consider two edge dislocations that have the same sign and the identical slip plane (Figure 7.5a).
Compressive and tensile strain fields for both lie on the same side of the slip plane; there exists between these two isolated dislocations a mutual repulsive force that tends to move them apart.
Dr. Mohammad Abuhaiba, PE

12. 7.3 Characteristics of Dislocations
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7.3 Characteristics of Dislocations
Two dislocations of opposite sign and having same slip plane will be attracted to one another (Figure 7.5b), the two extra half-planes of atoms will align and become a complete plane.
These strain fields and associated forces are important in the strengthening mechanisms for metals.
Dr. Mohammad Abuhaiba, PE

13. 7.4 Slip Systems Slip plane - plane allowing easiest slippage
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7.4 Slip Systems
Slip plane - plane allowing easiest slippage
Wide inter-planar spacing - highest planar densities
Slip direction - direction of movement - Highest linear densities
FCC Slip occurs on {111} planes in <110> directions
=> total of 12 slip systems in FCC
in BCC & HCP other slip systems occur
Dr. Mohammad Abuhaiba, PE

14. Table 7.1: Slip Systems for FCC, BCC, and HCP Metals
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Table 7.1: Slip Systems for FCC, BCC, and HCP Metals
Dr. Mohammad Abuhaiba, PE

15. 7.5 Slip in Single Crystals
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7.5 Slip in Single Crystals
• Crystals slip due to a resolved shear stress, tR.
• Applied tension can produce such a stress.
l = angle between stress direction & Slip direction
 = angle between normal to slip plane & stress direction
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
Dr. Mohammad Abuhaiba, PE

16. Critical Resolved Shear Stress
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Critical Resolved Shear Stress
• Condition for dislocation motion:
• Crystal orientation can make
it easy or hard to move dislocation
10-4 GPa to 10-2 GPa
typically tR
= 0 l
=90° s tR = s /2 l
=45° f tR
= 0 f
=90° s
Dr. Mohammad Abuhaiba, PE

17. 7.5 Slip in Single Crystals
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7.5 Slip in Single Crystals
Resolved  (shear stress) is maximum at  =  = 45º
And
CRSS = y/2 for dislocations to move (in single crystals)
Dr. Mohammad Abuhaiba, PE

18. Determining  and  angles for Slip in Crystals
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For cubic xtals, angles between directions are given by:
For metals we compare Slip System (normal to slip plane is a direction with exact indices as plane) to applied tensile direction using this equation to determine the values of  and  to plug into the R equation to determine if slip is expected
Dr. Mohammad Abuhaiba, PE

19. 4/22/2017 3:35 AM
EXAMPLE PROBLEM 7.1
Resolved Shear Stress and Stress-to-Initiate-Yielding Computations
Consider a single crystal of BCC iron oriented such that a tensile stress is applied along a [010] direction.
Compute the resolved shear stress along a (110) plane and in a direction when a tensile stress of 52 MPa is applied.
If slip occurs on a (110) plane and in a [-111] direction, and the critical resolved shear stress is 30MPa, calculate the magnitude of the applied tensile stress necessary to initiate yielding.
Dr. Mohammad Abuhaiba, PE

20. 7.6 Plastic Deformation of Polycrystalline Materials
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7.6 Plastic Deformation of Polycrystalline Materials s
300 mm
Stronger since grain boundaries pin deformations
Slip planes & directions (l, f) change from one crystal to another.
tR will vary from one crystal to another.
The crystal with the largest tR yields first.
Other (less favorably oriented) crystals yield (slip) later.
Dr. Mohammad Abuhaiba, PE

21. 7.6 Plastic Deformation of Polycrystalline Materials
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7.6 Plastic Deformation of Polycrystalline Materials
Ordinarily ductility is sacrificed when an alloy is strengthened.
The relationship between dislocation motion and mechanical behavior of metals is significance to the understanding of strengthening mechanisms.
The ability of a metal to plastically deform depends on the ability of dislocations to move.
Virtually all strengthening techniques rely on this simple principle: Restricting or Hindering dislocation motion renders a material harder and stronger.
We will consider strengthening single phase metals by:
grain size reduction
solid-solution alloying
strain hardening
Dr. Mohammad Abuhaiba, PE

22. 7.8 Strengthening by Grain Size Reduction
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7.8 Strengthening by Grain Size Reduction
Grain boundaries are barriers to slip.
Barrier "strength" increases with Increasing angle of mis-orientation.
Smaller grain size: more barriers to slip.
Dr. Mohammad Abuhaiba, PE

23. 7.8 Strengthening by Grain Size Reduction
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7.8 Strengthening by Grain Size Reduction
Hall-Petch equation:
Dr. Mohammad Abuhaiba, PE

24. 7.8 Strengthening by Grain Size Reduction
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7.8 Strengthening by Grain Size Reduction
Grain Size Reduction Techniques:
Increase Rate of solidification from the liquid phase.
Perform Plastic deformation followed by an appropriate heat treatment.
Notes:
Grain size reduction also improves toughness of many alloys.
Small-angle grain boundaries are not effective in interfering with the slip process because of the small crystallographic misalignment across the boundary.
Boundaries between two different phases are also impediments to movements of dislocations.
Dr. Mohammad Abuhaiba, PE

25. 7.9 Solid-solution Strengthening
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7.9 Solid-solution Strengthening
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
Dr. Mohammad Abuhaiba, PE

26. 7.9 Solid-solution Strengthening
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7.9 Solid-solution Strengthening
small impurities tend to concentrate at dislocations on the “Compressive stress side”
reduce mobility of dislocation  increase strength
Dr. Mohammad Abuhaiba, PE

27. 7.9 Solid-solution Strengthening
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7.9 Solid-solution Strengthening
Large impurities concentrate at dislocations on “Tensile Stress” side – pinning dislocation
Dr. Mohammad Abuhaiba, PE

28. 7.9 Solid-solution Strengthening
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7.9 Solid-solution Strengthening
• 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.
Dr. Mohammad Abuhaiba, PE

29. 7.10 Strain Hardening Room temperature deformation.
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7.10 Strain Hardening
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
Dr. Mohammad Abuhaiba, PE

30. 7.10 Strain Hardening Ti alloy after cold working:
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7.10 Strain Hardening
0.9 mm
Ti alloy after cold working:
Dislocations entangle and multiply
Thus, Dislocation motion becomes more difficult.
Dr. Mohammad Abuhaiba, PE

31. 7.10 Strain Hardening Dislocation density = total dislocation length
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7.10 Strain Hardening
Dislocation density =
Carefully grown single crystal
 ca. 103 mm-2
Deforming sample increases density
 mm-2
Heat treatment reduces density
 mm-2
total dislocation length
unit volume
large hardening
small hardening s e y0 y1
Again it propagates through til reaches the edge
• Yield stress increases
as rd increases:
Dr. Mohammad Abuhaiba, PE

32. 7.10 Strain Hardening As cold work is increased
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7.10 Strain Hardening
As cold work is increased
Yield strength (sy) increases.
Tensile strength (TS) increases.
Ductility (%EL or %AR) decreases.
Dr. Mohammad Abuhaiba, PE

33. Cold Work Analysis • What is the tensile strength &
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Cold Work Analysis
• What is the tensile strength &
ductility after cold working?
Cold
Work D d
=12.2mm
Copper D o
=15.2mm
Dr. Mohammad Abuhaiba, PE

34. Cold Work Analysis Cu Cu Cu
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Cold Work Analysis
What is the tensile strength & ductility after cold working to 35.6%?
% 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 7%
%EL
= 7%
340MPa
TS = 340MPa
YS = 300 MPa
Dr. Mohammad Abuhaiba, PE

35. s- e Behavior vs. Temperature
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s- e Behavior vs. Temperature
-200C
-100C
25C
800
600
400
200
Strain
Stress (MPa)
0.1
0.2
0.3
0.4
0.5
Results for polycrystalline iron:
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
Dr. Mohammad Abuhaiba, PE

36. Recovery, Recrystallization, and Grain Growth
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Recovery, Recrystallization, and Grain Growth
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
1 hour treatment at Tanneal...
decreases TS and increases %EL.
Effects of cold work are reversed!
3 Annealing stages to discuss...
Dr. Mohammad Abuhaiba, PE

37. 7.11 Recovery tR Annihilation reduces dislocation density.
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7.11 Recovery
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
Dr. Mohammad Abuhaiba, PE

38. 7.12 Recrystallization 0.6 mm New grains are formed that:
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7.12 Recrystallization
New grains are formed that:
have a low dislocation density
are small
consume cold-worked grains.
33% cold
worked
brass
New crystals
nucleate after
3 sec. at 580C.
0.6 mm
Dr. Mohammad Abuhaiba, PE

39. 7.12 Recrystallization • All cold-worked grains are consumed. 0.6 mm
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7.12 Recrystallization
• All cold-worked grains are consumed.
After 4
seconds
After 8
0.6 mm
Dr. Mohammad Abuhaiba, PE

40. 4/22/2017 3:35 AM
7.12 Recrystallization
Dr. Mohammad Abuhaiba, PE

41. TR = recrystallization temperature
4/22/2017 3:35 AM TR
TR = recrystallization temperature
Dr. Mohammad Abuhaiba, PE

42. 4/22/2017 3:35 AM
DESIGN EXAMPLE 7.1
Description of Diameter Reduction Procedure A cylindrical rod of non-cold-worked brass having an initial diameter of 6.4 mm is to be cold worked by drawing such that the cross-sectional area is reduced. It is required to have a cold-worked yield strength of at least 345 MPa and a ductility in excess of 20%EL; in addition, a final diameter of 5.1 mm is necessary. Describe the manner in which this procedure may be carried out.
Dr. Mohammad Abuhaiba, PE

43. 7.13 Grain Growth At longer times, larger grains consume smaller ones.
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7.13 Grain Growth
After 8 s,
580ºC
After 15 min,
0.6 mm
At longer times, larger grains consume smaller ones.
Why? Grain boundary area (and therefore energy) is reduced.
Empirical Relation:
coefficient dependent on material & Temp.
elapsed time
exponent typ. ~ 2
grain dia. At time t.
Dr. Mohammad Abuhaiba, PE
This is: Ostwald Ripening

44. Recrystallization Temperature, TR
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Recrystallization Temperature, TR
TR = recrystallization temperature = point of highest rate of property change
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
Dr. Mohammad Abuhaiba, PE

45. Coldwork Calculations
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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.
Dr. Mohammad Abuhaiba, PE

46. Coldwork Calculations Solution
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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
Dr. Mohammad Abuhaiba, PE

47. Coldwork Calc Solution: Cont.
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Coldwork Calc Solution: Cont.
420
540 6
For %CW = 43.8%
y = 420 MPa
TS = 540 MPa > 380 MPa
%EL = 6 < 15
Dr. Mohammad Abuhaiba, PE
This doesn’t satisfy criteria…… what can we do?

48. Coldwork Calc Solution: Cont.
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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%
Dr. Mohammad Abuhaiba, PE

49. This process Needs an Intermediate Recrystallization
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This process Needs an Intermediate Recrystallization
Dr. Mohammad Abuhaiba, PE
Cold draw-anneal-cold draw again
For objective we need a cold work of %CW  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 =

50. Coldwork Calculations Solution
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Coldwork Calculations Solution
Summary:
Cold work D01= 0.40 in  Df1 = in
Anneal above Ds2 = Df1
Cold work Ds2= in  Df 2 =0.30 in
Therefore, meets all requirements 
Fig 7.19
Dr. Mohammad Abuhaiba, PE

51. Summary Dislocations are observed primarily in metals and alloys.
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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 .
Dr. Mohammad Abuhaiba, PE

52. HomeWork Assignment 7, 12, 17, 24, 29, 41, D6 Due Monday 21/11/2011
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HomeWork Assignment
7, 12, 17, 24, 29, 41, D6
Due Monday 21/11/2011
Dr. Mohammad Abuhaiba, PE