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. Deformation Mechanisms
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 (relatively close-packed) in <110> directions (close-packed)
=> total of 12 slip systems in FCC
in BCC & HCP other slip systems occur
Adapted from Fig. 7.6, Callister 7e.

3. Slip Systems that can and will operate in the Cubic Metals:

4. 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
Note: By definition  is the angle between the stress direction and Slip direction;  is the angle between the normal to slip plane and stress direction

5. 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

6. Generally:
Resolved  (shear stress) is maximum at  =  = 45º
And
CRSS = y/2 for dislocations to move (in single crystals)

7. Determining  and  angles for Slip in Crystals (single X-tals this is easy!)
 and  angles are respectively angle between tensile direction and Normal to Slip plane and angle between tensile direction and slip direction (these slip directions are material dependent)
And Remembering for cubic xtals, angles between directions are given by:
Thus 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

8. Slip Motion in Polycrystals
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.
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].)

9. After seeing the effect of poly crystalline
After seeing the effect of poly crystalline materials we can say (as related to strength):
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, and strain hardening

10. Strategies for Strengthening: 1: Reduce Grain Size
• Grain boundaries are
barriers to slip.
• Barrier "strength"
increases with
Increasing 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.)

11. Hall-Petch equation:

12. 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.

13. 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

14. Stress Concentration at Dislocations
Don’t move past one another – hardens material
Adapted from Fig. 7.4, Callister 7e.

15. Strengthening by Alloying
small impurities tend to concentrate at dislocations on the “Compressive stress side”
reduce mobility of dislocation  increase strength
Adapted from Fig. 7.17, Callister 7e.

16. Strengthening by alloying
Large impurities concentrate at dislocations on “Tensile Stress” side – pinning dislocation
Adapted from Fig. 7.18, Callister 7e.

17. 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:
Adapted from Fig (a) and (b), Callister 7e.
• Alloying increases sy and TS.

18. Strategies for Strengthening: 3. Precipitation Strengthening
• Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum).
Side View
precipitate
Top View
Unslipped part of slip plane S
Large shear stress needed to move dislocation toward precipitate and shear it.
Dislocation “advances” but precipitates act as “pinning” sites with spacing S. which “multiplies” Dislocation density
Slipped part of slip plane
• Result:

19. 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 & H.T.
1.5mm
Adapted from Fig , Callister 7e.
(Fig is courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

20. 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
-Rolling
roll A o d
-Drawing
tensile
force A o d
die
-Extrusion
ram
billet
container
force
die holder
die A o d
extrusion

21. During Cold Work • Ti alloy after cold working:
0.9 mm
• Dislocations entangle and multiply
• Thus, Dislocation motion becomes more difficult.
Adapted from Fig. 4.6, Callister 7e.
(Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.)

22. Result of Cold Work Dislocation density = total dislocation length
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
Again it propagates through til reaches the edge
large hardening
small hardening s e y0 y1
• Yield stress increases
as rd increases:

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

24. Cold Work Analysis • What is the tensile strength &
ductility after cold working?
Cold
Work D d
=12.2mm
Copper D o
=15.2mm

25. Cold Work Analysis Cu Cu Cu • 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
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.)

26. 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

27. 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...
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.)

28. 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

29. 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
Adapted from Fig (a),(b), Callister 7e. (Fig (a),(b) are courtesy of J.E. Burke, General Electric Company.)

30. 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.)

31. 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

32. 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
Adapted from Fig (d),(e), Callister 7e. (Fig (d),(e) are courtesy of J.E. Burke, General Electric Company.)
coefficient dependent on material & Temp.
• Empirical Relation:
elapsed time
exponent typ. ~ 2
grain dia. At time t.
This is: Ostwald Ripening

33. TR = recrystallization temperatureTR
TR = recrystallization temperature
Adapted from Fig. 7.22, Callister 7e.

34. 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.

35. 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

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

37. Coldwork Calc Solution: 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%

38. This process Needs an Intermediate Recrystallization
i.e.: 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 =

39. 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

40. 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.