Material Science ME 3101 Fall 2016 ME 3101

1. Chapter 7 - 1
Chapter 7:
Dislocations & Strengthening
Mechanisms
Dr. Mir Atiqullah
Chapter 7 - 2
Ch7: Do we know these Terms?
• Cold Working
• Dislocation Density
• Grain growth-
• Crystal lattice
• Re-crystallization
• Recovery
• Slip
• Strain hardening
• Solid Solution strengthening
• Anisotropy
• Resolved Shear Stress
Read Thru end of the
chapter 7 specially
mechanisms of
strengthening
Study example
problems-
Practice problems from
end of chapter.
Chapter 7 - 3
Dislocations & Materials Classes
• Covalent Ceramics
(Si, diamond): Motion hard.
-directional (angular) bonding
• Ionic Ceramics (NaCl):
Motion hard.
-need to avoid ++ and - - neighbors.
Which arrow shows best slip direction?
+ + + +
+++
+ + + +
- - -
----
- - -
• Metals: Disl. motion easier.
-non-directional bonding
-close-packed directions
for slip.
electron cloud ion cores
+
+
+
+
+++++++
+ + + + + +
+++++++
Polymers?
Chapter 7 - 4
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.

2. Chapter 7 -
You tube for you..dislocation
• https://www.youtube.com/watch?v=quGSf
mt4V6Y
• https://www.google.com/search?q=video+
dislocation+motion&sourceid=ie7&rls=co
m.microsoft:en-US:IE-
Address&ie=&oe=&gws_rd=ssl
5 Chapter 7 - 6
Dislocation Motion-Edge , screw
• Dislocation moves along slip plane in slip direction –
WHY?
• perpendicular to dislocation line
• Slip direction same direction as Burgers vector
Edge dislocation
Screw dislocation
Adapted from Fig. 7.2,
Callister 7e.
Chapter 7 - 7
Dislocation Characteristics –
local stress, annihilation, repulsion,..
Stress due to an edge
dislocation strain fields
Interaction between dislocations-
Dislocation annihilation.
Plastic Deformation- causes dramatic increase in dislocation-
upto 1010/mm2 Source: existing dislocations, grain boundary,
internal defects, surface irregularities etc.
Chapter 7 - 8
Slip System (in planes along some directions)
– Slip plane - plane allowing easiest slippage
• 1-Wide interplanar spacings – 2-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
Deformation Mechanisms
Adapted from Fig.
7.6, Callister 7e.

3. Chapter 7 - 9
7.5 Stress and Dislocation Motion (Slip)
• Crystals slip due to a resolved shear stress, R.
• Applied tension can produce such a stress.
slip plane
normal, ns
Resolved shear
stress: R =Fs/As
AS
R
R
FS
Relation between
 and R
R=FS/AS
Fcos  A/cos 

F
FS
nS
AS
A
Applied tensile
stress: = F/A
F
A
F
 coscosR
Chapter 7 - 10
• Condition for dislocation motion: CRSSR
• Crystal orientation can make
it easy or hard to move dislocation 10-4 GPa to 10-2 GPa
A property, typically
 coscosR
Critical Resolved Shear Stress (CRSS)
 maximum at  =  = 45º
R = 0
=90°

R = /2
=45°
 =45°

R = 0
=90°

Chapter 7 - 11
Macroscopic Slip -
Single Crystal
Microscopic slip in single crystal
Zn Single
Crystal
Chapter 7 - 12
Ex: Deformation of single crystal
the applied stress of 6500 psi will
not cause the crystal to yield.
   cos  cos 
  6500 psi
=35°
=60°

  (6500 psi) ( cos 35
)(cos 60
)
 (6500 psi) ( 0.41)
  2662 psi  crss  3000 psi
crss = 3000 psi
a) Will the single crystal yield?
b) If not, what stress is needed?
 = 6500 psi
Adapted from
Fig. 7.7,
Callister 7e.
So ?

4. Chapter 7 - 13
Ex: Deformation of single crystal
psi7325
41.0
psi3000
coscos
crss 


 y
Then What stress is necessary to yield
(i.e., what is the yield stress, y)?
)41.0(coscospsi3000crss yy 
psi7325 y
So for deformation to occur the applied stress must
be greater than or equal to the yield stress
Chapter 7 - 14
• Stronger - grain boundaries
pin deformations
• 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.
( yield Jogging?)
• Other (less favorably
oriented) crystals
yield later.
Adapted from Fig.
7.10, Callister 7e.
(Fig. 7.10 is courtesy
of C. Brady, National
Bureau of Standards
[now the National
Institute of Standards
and Technology,
Gaithersburg, MD].)
Slip Motion in Polycrystals
The picture can't be displayed.

300
m
Chapter 7 - 15
• Can be induced by rolling a polycrystalline metal
- before rolling
235 m
- isotropic
since grains are
approx. spherical
& randomly
oriented.
- after rolling
- anisotropic
since rolling affects grain
orientation and shape.
rolling direction
Adapted from Fig. 7.11,
Callister 7e. (Fig. 7.11 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.)
Anisotropy
Chapter 7 - 16
Technical Term English
1. Reduce Grain Size (grain refinement)
2. Solid Solution (alloying) (atomic level)
3. Precipitation hardening (grain level)
4. Cold working – (work hardening)
4 Strategies for Strengthening:

5. Chapter 7 - 17
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: (p-189)
21/
yoyield dk 

Adapted from Fig. 7.14, Callister 7e.
(Fig. 7.14 is from A Textbook of Materials
Technology, by Van Vlack, Pearson Education,
Inc., Upper Saddle River, NJ.)
St. line on a semi-log plot Fig 7.15 p-189
Chapter 7 - 18
Example Problem:
Strengthening by Reducing Grain Size
7.22** Given grain Diam. d1= 5E-2 mm, y1=135 Mpa
and at grain Diam. d2= 8E-3 mm y2=260 Mpa
To achieve y= 205 Mpa what should be grain size; d=?
Solution Hints:
1. First determine the co-efficients of Hall-Petch Eqtn.
How: 2 unknowns, use two sets of given data (2
equations).
1. o= ? Ky = ?
2. Then use these and y= 205 Mpa to calc. d = ?
yield  o  kyd1/2
Can we custom strengthen metals
by controlling grain size.
Chapter 7 - 19
• Impurity atoms distort the lattice & generate stress.
• Stress can produce a barrier to dislocation motion.
Strategies for Strengthening:
2: Solid Solutions
• 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
Chapter 7 - 20
Stress Concentration at Dislocations
Adapted from Fig. 7.4,
Callister 7e.

6. Chapter 7 - 21
Strengthening by Alloying-how?
• small impurities tend to concentrate at dislocations
• reduce mobility of dislocation  increase strength
Adapted from Fig.
7.17, Callister 7e.
Chapter 7 - 22
Strengthening by alloying
• large impurities concentrate at dislocations on low
density side
Adapted from Fig.
7.18, Callister 7e.
Chapter 7 - 23
Ex: Solid Solution
Strengthening in Copper – experimental
• Tensile strength & yield strength increase with wt% Ni.
• Empirical relation:
• Alloying increases y and TS.
21/
y C~
Adapted from Fig.
7.16 (a) and (b),
Callister 7e.
Tensilestrength(MPa)
wt.% Ni, (Concentration C)
200
300
400
0 10 20 30 40 50
Yieldstrength(MPa)
wt.%Ni, (Concentration C)
60
120
180
0 10 20 30 40 50
Chapter 7 - 24
• Hard precipitates are difficult to shear.
Ex: Ceramics in metals (SiC in Iron or Aluminum).
• Result:
S
~y
1

Strategies for Strengthening:
3: Precipitation Strengthening
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
S spacing
From superstuarted solid solution (solution heat treatment) by quenching
and then by precipitation (natural aging at room temperature or artificial
aging at an elevated temperature). E.g Al-Cu, Cu-Be, Cu-Sn, Mg-Al.

7. Chapter 7 - 25
• Internal wing structure on Boeing 767
• Aluminum is strengthened with precipitates formed
by alloying.
Adapted from Fig.
11.26, Callister 7e.
(Fig. 11.26 is courtesy
of G.H. Narayanan
and A.G. Miller,
Boeing Commercial
Airplane Company.)
1.5m
Application:
Precipitation Strengthening
Adapted from chapter-
opening photograph,
Chapter 11, Callister 5e.
(courtesy of G.H.
Narayanan and A.G.
Miller, Boeing Commercial
Airplane Company.)
Chapter 7 - 26
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
Ao Ad
force
die
blank
force-Drawing
tensile
force
Ao
Addie
die
-Extrusion
ram billet
container
container
force
die holder
die
Ao
Adextrusion
100x%
o
do
A
AA
CW


-Rolling
roll
Ao
Ad
roll
Chapter 7 - 27
• Ti alloy after cold working:
• 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.)
Dislocations During Cold Work
0.9 m
Chapter 7 - 28
Result of Cold Work
Dislocation density =
– Carefully grown single crystal
 ca. 103 mm-2
– Deforming sample increases density
 109-1010 mm-2
– Heat treatment reduces density
 105-106 mm-2
• Yield stress increases
as d increases:
total dislocation length
unit volume
large hardening
small hardening


y0
y1

8. Chapter 7 - 29
Effects of Stress at Dislocations
Adapted from Fig.
7.5, Callister 7e.
Chapter 7 - 30
Impact of Cold Work
Adapted from Fig. 7.20,
Low carbon steel
• Yield strength (y) increases.
• Tensile strength (TS) increases.
• Ductility (%EL or %AR) decreases.
As cold work is increased
Chapter 7 - 31
• Expl7.2: What is the tensile strength &
ductility after cold working?
Adapted from Fig. 7.19, Callister 7e. (Fig. 7.19 is adapted from Metals Handbook)
%6.35100x%
2
22




o
do
r
rr
CW
Cold Work Analysis
% Cold Work
100
300
500
700
Cu
200 40 60
yield strength (MPa)
y = 300MPa
300MPa
% Cold Work
tensile strength (MPa)
200
Cu
0
400
600
800
20 40 60
ductility (%EL)
% Cold Work
20
40
60
20 40 6000
Cu
Do =15.2mm
Cold
Work
Dd =12.2mm
Copper
340MPa
TS = 340MPa
7%
%EL = 7%
Chapter 7 - 32
An Example Problem
7.28** Two previously undeformed cylindrical specimens of an alloy are to be strain
hardened by reducing their cross-sectional areas (maintaining circular shape).
Specimen 1: initial and final radii are 15 mm and 12 mm.
Specimen 2: initial radius is 11 mm, and must have the same deformed hardness as
the first one.
Calc. Final radius of the specimen 2.
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
%CW =
Ao  Ad
Ao
x 100 =
 ro
2
 rd
2
r
o
2
x 100
=
 (15 mm)
2
 (12 mm)
2
 (15 mm)2
x 100 = 36%CW
rd = ro 1 
%CW
100
= (11 mm) 1 
36%CW
100
= 8.80 mm
For the second specimen, the deformed radius is computed using the same
% CW
equation and solving for rd as

9. Chapter 7 - 33
• Results for
polycrystalline iron:
• y and TS decrease with increasing test temperature.
• %EL increases with increasing test temperature.
• Why? Vacancies
help dislocations
move past obstacles.
Adapted from Fig. 6.14,
Callister 7e.
- Behavior vs. Temperature
2. vacancies
replace
atoms on the
disl. half
plane
3. disl. glides past obstacle
-200C
-100C
25C
800
600
400
200
0
Strain
Stress(MPa)
0 0.1 0.2 0.3 0.4 0.5
1. disl. trapped
by obstacle
obstacle
Chapter 7 - 34
TR
Adapted from Fig. 7.22,
Callister 7e.
º
º
TR = recrystallization
temperature
What are the 3
Annealing stages?
1.
2.
3.
Chapter 7 - 35
Annihilation reduces dislocation density.
Recovery
• Scenario 1
Results from
diffusion
• Scenario 2
4. opposite dislocations
meet and annihilate
Dislocations
annihilate
and form
a perfect
atomic
plane.
extra half-plane
of atoms
extra half-plane
of atoms
atoms
diffuse
to regions
of tension
2. grey atoms leave by
vacancy diffusion
allowing disl. to “climb”
R
1. dislocation blocked;
can’t move to the right
Obstacle dislocation
3. “Climbed” disl. can now
move on new slip plane
Chapter 7 - 36
• New grains are formed that:
-- have a small dislocation density
-- are small
-- consume cold-worked grains.
Adapted from
Fig. 7.21 (a),(b),
Callister 7e.
(Fig. 7.21 (a),(b)
are courtesy of
J.E. Burke,
General Electric
Company.)
33% cold
worked
brass
New crystals
nucleate after
3 sec. at 580C.
0.6 mm 0.6 mm
Recrystallization

10. Chapter 7 - 37
• All cold-worked grains are consumed.
Adapted from
Fig. 7.21 (c),(d),
Callister 7e.
(Fig. 7.21 (c),(d)
are courtesy of
J.E. Burke,
General Electric
Company.)
After 4
seconds
After 8
seconds
0.6 mm0.6 mm
Further Recrystallization
Chapter 7 - 38
• At longer times, larger grains consume smaller ones.
• Why? Grain boundary area (and therefore energy)
is reduced.
After 8 s,
580ºC
After 15 min,
580ºC
0.6 mm 0.6 mm
Adapted from
Fig. 7.21 (d),(e),
Callister 7e.
(Fig. 7.21 (d),(e)
are courtesy of
J.E. Burke,
General Electric
Company.)
Grain Growth
• Empirical Relation:
Ktdd n
o
n

elapsed time
coefficient dependent
on material and T.
grain diam.
at time t.
exponent typ. ~ 2
Ostwald Ripening
Chapter 7 - 39
Recrystallization Temperature, TR
TR = recrystallization temperature = point of
highest rate of property change-depends on
phase diagram of the alloy. Generally:
1. TR  0.3-0.6 Tm (K) - Tm melting temp.
2. Due to diffusion  annealing time (t) TR = f(t)
shorter annealing time => need higher TR
3. Higher %CW => lower TR – strain hardening
4. Pure metals lower TR due to dislocation
movements
• Easier to move in pure metals. So lower TR
Chapter 7 - 40
7.D6 CW Calculations
Problem:
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).
Q: How can this be done?

11. Chapter 7 - 41
CW Calc. Solution
If we directly draw to the final diameter
what happens?
%843100x
400
300
1100x
4
4
1
1001100x%
2
2
2
.
.
.
D
D
x
A
A
A
AA
CW
o
f
o
f
o
fo



































 

Do = 0.40 in
Brass
Cold
Work
Df = 0.30 in
Chapter 7 - 42
CW Calc. Solution: Cont.
• For %CW = 43.8%
Adapted from Fig.
7.19, Callister 7e.
540420
– y = 420 MPa
– TS = 540 MPa > 380 MPa
6
– %EL = 6 < 15
• This doesn’t satisfy criteria…… what can we do?
Chapter 7 - 43
Coldwork Calc Solution: Cont.
Adapted from Fig.
7.19, Callister 7e.
380
12
15
27
For %EL > 15
For TS > 380 MPa > 12 %CW
< 27 %CW
 our working range is limited to %CW = 12-27
Chapter 7 - 44
CW Calc. Soln: Recrystallization
Cold draw-anneal-cold draw again
• For objective we need a cold work of %CW  12-27
– Let’s try %CW = 20
• Diameter after first cold draw (before 2nd cold draw)?
– must be calculated as follows:
100
%
11001% 2
02
2
2
2
02
2
2 CW
D
D
x
D
D
CW ff









50
02
2
100
%
1
.
f CW
D
D






 50
2
02
100
%
1
.
f
CW
D
D









m3350
100
20
1300
50
021 ..DD
.
f 





Intermediate diameter =

12. Chapter 7 - 45
Coldwork Calculations Solution
Summary:
1. Cold work D01= 0.40 in  Df1 = 0.335 m
2. Anneal above D02 = Df1
3. Cold work D02= 0.335 in  Df2 =0.30 m
Therefore, meets all requirements
20100
3350
30
1%
2
2 














 x
.
.
CW
24%
MPa400
MPa340



EL
TS
y

%CW1  1
0.335
0.4






2







x 100  30
Fig 7.19
Chapter 7 - 46
Grain Growth ( comments)
• Driving force- Decrease in energy with larger
diameters ( lower surface area)
• Occurs by migration of boundaries – just like
political/national powers : larger grains grows at the
expense of smaller ones
• Atoms ‘pop’ in opposite direction to boundary
migration direction.
• Polycrystalline material: Ktdd nn
 0
Chapter 7 - 47
Do we know these now?
• Cold Working : Measured by ____?
• Dislocation Density
• Grain growth- Single most factor is _______?
• Crystal lattice
• Re-crystallization.
• Recovery
• Slip
• Strain hardening
• Solid Solution strengthening
• Anisotropy. Examples of anisotropic material….?
• CRSS ? ______ ______ Shear Stress. Formula?
• Resolved Shear Stress: Formula?
Chapter 7 - 48
• 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.
Summary