Lecture 09

Dislocations & Strengthening Mechanisms

Chapter 7 - 1
 Dislocations & Strengthening Mechanisms

ISSUES TO ADDRESS...
• Why are dislocations observed primarily in metals
and alloys?

g and dislocation motion related?

• How are strength

• How do we increase strength?

• How can heating change strength and other properties?

Chapter 7 - 2
 Mechanical Behavior & Structure

Controlled by Controlled by Processing

Chemical bond via defects, crystal structure
“Chemistry” andd microstricture
i i
Chapter 7 - 3
 Dislocations & Materials Classes
• Metals: Disl. motion easier.
+ + + + + + + +
-non-directional bonding + + + + + + + +
-close-packed directions + + + + + + + +
for slip. electron cloud ion cores

• Covalent Ceramics
(Si, diamond): Motion hard.
-directional
directional (angular) bonding

• Ionic Ceramics (NaCl):

+ - + - + - +
Motion hard.
- + - + - + -
-need to avoid ++ and - -
g
neighbors. + - + - + - +

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!

Chapter 7 - 5
 Dislocation Motion (Cont’d)
Caterpillar…


b

Chapter 7 - 6
 Dislocation Motion
• Dislocation moves along slip plane in slip direction
perpendicular
p p to dislocation line
• Slip direction same direction as Burgers vector
Edge dislocation

Screw dislocation

Chapter 7 - 7
Comparison of Dislocation Motion
“Edge vs Screw Dislocations”

Chapter 7 - 8
 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 (close-packed) in <110>

directions (close-packed) => total of 12 slip systems in FCC
– in BCC & HCP other slip systems occur
Chapter 7 - 9
Deformation Mechanisms
“Common Slip Systems”

Chapter 7 - 10
Slip Systems in Common Metals

After Dieter,
Mechanical Metallurgy
(1990)

Chapter 7 - 11
 Stress and Dislocation Motion
• Crystals slip due to a resolved shear stress, tR.
• Applied tension can produce such a stress.
Applied tensile Resolved shear Relation between
= s F/A
stress: stress: tR =Fs /A s s and tR
F slip plane tR = FS /AS
A normal, ns
tR
AS Fcos l A/cos f
FS
F nS f
l A
tR FS AS
F
Schmid’s Law
R   cos  cos 
Chapter 7 - 12
 Critical Resolved Shear Stress
• Condition for dislocation motion: R  CRSS
• Crystal orientation can make typically
it easy or hard to move dislocation
10-4 GPa to 10-2 GPa
R   cos  cos 
s s s

tR = 0 tR = s /2 tR = 0
l =90°
90° l =45°
45° f =90°
90°
f =45°

 maximum at  =  = 45º Chapter 7 - 13

Single Crystal Slip
Single
crystal
wire
i

Chapter 7 - 14
 Ex: Deformation of single crystal
a) Will the single crystal yield?
b) If not, what stress is needed?
=60°

crss = 3000 psii
=35°
   cos  cos 
  6500 psi

  (6500 psi) (cos 35 )(cos 60 )

 (6500 psi) (0.41)
  2662 psi  crss  3000 psi
 = 6500 psi
So the applied stress of 6500 psi will not cause the crystal
to yield. Chapter 7 - 15
 Ex 1: Deformation of single crystal
What stress is necessary (i.e., what is the yield stress, y)?

crss  3000 psi   y cos  cos    y (0.41)

crss 3000 psi

 y    7325 psi
cos  cos  0.41

So for deformation to occur the applied stress must be

greater than or equal to the yield stress

   y  7325 psi
Chapter 7 - 16
 Ex 2: Deformation of Single crystal

After Dieter, Mechanical Metallurgy (1990)

Chapter 7 - 17
 Slip Motion in Polycrystals
• Stronger - grain boundaries
pin deformations 
• Slip planes & directions
change from one
crystal
t l to
t another.
th
• R will vary from one
crystal to another.
another
• The crystal with the
largest R yields first
first.
• Other (less favorably
oriented) crystals
300 mm
yield later.

Chapter 7 - 18
 Anisotropy in y
• Can be induced by rolling a polycrystalline metal
- before rollingg - after rollingg

rolling direction
235 mm
- isotropic - anisotropic
since grains are since rolling affects grain
approx. spherical orientation and shape.
& randomly
oriented.

Chapter 7 - 19
 Anisotropy in Deformation
1. Cylinder of 2. Fire cylinder 3. Deformed
Tantalum at a target. cylinder
machined
from a
rolled plate: side view
direction
rolling d

end plate
thickness
view direction
• The noncircular end view shows
anisotropic deformation of rolled material.
Chapter 7 - 20
 Deformation by Mech’l Twinning
Twins also re-orient slip planes and contribute to dislocation slip indirectly…

Twinning is an alternate (plastic)

deformation mechanism –observed when the
strain rate is very high or at low T (dislocation slip is suppressed)
It is a strain-relief mechanism. Very common in BCC and HCP metals
when slip is restricted! Chapter 7 -
Mech’l Twinning vs. Dislocation Slip

Homogeneous
(Twin Band)

After
Dieter.
Mech’l Metallurgy Chapter 7 -
 Strategies for Strengthening:
11. Reduce
R d Grain
G i Size
Si

• Grain boundaries are

barriers to slip.
• Barrier "strength"
strength
increases with
increasing angle of
misorientation.
i i i
• Smaller grain size (d):
p
more barriers to slip.

1/ 2
• Hall-Petch Equation:  yield  o  k y d
Chapter 7 -
 Strategies for Strengthening:
1.
1 RReduce
d Grain
G i Size (Cont’d)
Si (C t’d)

Very difficult
to synthesize polycrystalline
materials with d<100 nm

1/ 2
 yield
i ld   o  k y d

Chapter 7 -
 Strategies for Strengthening:
1.
1 Reduce Grain Size (Cont’d)
Primary mechanism
for strengthening by grain
boundaries…

Chapter 7 - 25
 Strategies for Strengthening:
22. S
Solid Solutions
lid S l ti
• Impurity atoms distort the lattice & generate stress.
• Stress can produce a barrier to dislocation motion.
• Smaller substitutional impurity • Larger substitutional impurity

A C

B D

Impurity generates local stress at A Impurity generates local stress at C

and B that opposes
pp dislocation motion and D that opposes
pp dislocation motion
to the right. to the right.

Chapter 7 - 26
Stress Concentration at Dislocations
“What makes a dislocation move & interact?”

Th stress field
The fi ld Around
A d the
h di
dislocation
l i
interacts with the applied stress (via R)
and the Strain field of structural
p f
imperfections such as grain
g boundaries
and solute atoms…
Chapter 7 -
 Strengthening by Alloying
• Small impurities tend to concentrate at dislocations
• Reduce mobility of dislocation;  increase strength

Chapter 7 - 28
 Strengthening by alloying (Cont’d)
• Large impurities concentrate at dislocations on low
densityy side

Chapter 7 -
 Ex: Solid Solution Strengthening in Copper

• Tensile strength & yield strength increase with wt% Ni.

180
400
Tensile sttrength (MPa)

Pa)
Yield strength (MP
120
300

200 60
0 10 20 30 40 50 0 10 20 30 40 50
wt % Ni
wt.% Ni, (Concentration C) wt %Ni (Concentration C)
wt.%Ni,

• Empirical relation: y ~ C1/ 2

• Alloying increases sy and TS.
Chapter 7 -
Dislocation Sources: Frank-
Frank-Read Sources

Chapter 7 - 31
 Strategies for Strengthening:
3.
3 Precipitation
P i it ti Strengthening
St th i
• Hard precipitates are difficult to shear.
Ex: Ceramics in metals (SiC in Iron or Aluminum).
precipitate Large shear stress needed
to move dislocation toward
Side View precipitate
i it t andd shear
h it.
it

pp p
Unslipped pp
part of slip plane
T View
Top Vi Dislocation
“advances” but
precipitates act as
S “pinning” sites with
spacing
p g S.
Slipped part of slip plane

The smaller the S, the larger the

1
• Result: y ~ dislocation line needs to bow out,,
S the more surface energy needs to
be spend!
Chapter 7 -
 Application:: Precipitation
Application Precipitation Strengthening
• Internal wing structure on Boeing 767

• Aluminum is strengthened with precipitates formed

b alloying.
by ll i

Black spots
p
are the ppts.

1.5mm
Chapter 7 - 33
 Strategies for Strengthening:
44. Cold
C ld Work (%CW))
W k (%CW
• Room temperature deformation.
• Common forming operations change the cross sectional area:

-Forging force -Rolling

roll
die Ad
A o blank Ad Ao
roll
force
-Drawing -Extrusion
Ao
die Ad container die holder
Ao tensile force
force ram billet extrusion Ad
die container die
Ao  Ad
%CW  x 100
Ao
Chapter 7 -
 Dislocations During Cold Work

• Ti alloy after cold working:

• Dislocations entangle with one

another during cold work.

• Dislocation motion becomes

more difficult.

 Strain hardening due to

dislocation-dislocation
interactions
cutting through a forest of
“cutting
0 9 mm
0.9 di l
dislocations…”
dislocations…
i ”
Chapter 7 -
Dislocations During Cold Work
Edge and screw
dislocation interaction
leads to strain hardening

After
Dieter.
Mech’l Metallurgy
gy

Chapter 7 - 36
 Result of Cold Work
total dislocation length
Dislocation density =
unit volume
– Carefully grown single crystal  ca. 103 mm-2
– Deforming sample increases density 109-1010 mm-2
– Heat
H t treatment
t t t reduces
d d it  105-10
density 106 mm-22

• Yield stress increases 

as rd increases:
y large hardening
1
y small hardening
0


Chapter 7 -
Effects of Stress at Dislocations

Chapter 7 -
 Impact of Cold Work
As cold work is increased:
• Yield strength (y) increases.
• Tensile strength (TS) increases.
• Ductility (%EL or %AR) decreases.

Chapter 7 -
 Cold Work Analysis
• What is the tensile strength & Copper
ductility after cold working? Cold
Work
ro2  rd2
%CW  x 100  35.6%
2
ro Do =15.2mm Dd =12.2mm

yield strength tensile strength ductility

(MPa) (MPa) 60 (%EL)
700 800

500 600 40

300
300MPa Cu
Cu 400 340MPa 20
Cu 7%
100
0 20 40 60 200 00
0 20 40 60 20 40 60
% Cold Work % Cold Work % Cold Work
y = 300MPa TS = 340MPa
340MP %EL = 7%

Chapter 7 -
 -  Behavior vs. Temperature
• Results for 800
-200C

MPa)
polycrystalline iron: 600

Stress (M
-100C
100C
400

200 25C

0
0 0.1 0.2 0.3 0.4 0.5
Strain
• y and TS decrease with increasing g test temperature.
p
• %EL increases with increasing test temperature.
3 . disl. glides past obstacle
• Why? Vacancies
p dislocations
help 2. vacancies
move past obstacles. replace
atoms on the
obstacle
disl. half
plane 1. disl. trapped
b obstacle
by bt l

Chapter 7 -
 %CW
Effect of Heating After %CW
• 1 hour treatment at Tanneal…decreases TS & increases %EL.
• Effects of cold work are reversed!
Annealing temperature (ºC)
100 200 300 400 500 600 700
600 60
gth (MPa))

tensile strength

ductility ((%EL)
50
500
3 Annealing
tenssile streng

40
stages to discuss...
discuss...
400 30

ductility 20
300

Chapter 7 -
 Recovery
Annihilation reduces dislocation density.
• Scenario 1 extra half-plane
of atoms Dislocations
Results from annihilate
diffusion atoms
diffuse and form
to regions a perfect
atomic
of tension
extra half-plane plane.
of atoms
• Scenario
S i 2
3 . “Climbed” disl. can now tR
move on new slip plane
2 . grey atoms leave by
4. opposite dislocations
vacancy diffusion
meet and annihilate
allowing disl. to “climb”
1. dislocation blocked; Obstacle dislocation
can’t move to the right

Chapter 7 -
 Recrystallization
• New grains are formed that:
- Have a small dislocation density
- Are
A small ll
- Consume cold-worked grains.
0 6 mm
0.6 0 6 mm
0.6

33% cold New crystals

worked
k d nucleate
l after
f
brass 3 sec. at 580C.
Chapter 7 -
 Further Recrystallization
• All cold
cold--worked grains are consumed.

0.6 mm 0.6 mm

After 4 After 8
seconds
d d
seconds

Chapter 7 -
 Grain Growth
• At longer times, larger grains consume smaller ones.
• Why? Grain boundary area (and therefore energy) is reduced.

0.6 mm 0.6 mm

After 8 s, After 15 min,

580ºC 580ºC

Coefficient dependent
• Empirical Relation:
on material and T.
Eexponent typ. ~ 2
Grain diam. Elapsed time
at time t. d n
 d on  Kt
Ostwald Ripening
Chapter 7 -
 º

Grain Growth

TR = Recrystallization
temperature

TR

Cold work state, is

A metastable state.
The energy stored
Due to cold work
is the driving for
Recrystallization.
Recrystallization
Recrystallized
grain are strain
free.
º
Chapter 7 -
 Recrystallization Temperature, TR
TR = recrystallization temperature = point of
highest rate of property change

11. Tm => TR  0.3-0.6

0 3 0 6 Tm (K)
2. Due to diffusion  annealing time TR = f(t)
shorter annealing time => higher TR
3. Higher %CW => lower TR – strain hardening
4. Pure metals lower TR due to dislocation movements
Easier to move in pure metals => lower TR

Chapter 7 -
Recrystallization Temperature, TR

Chapter 7 - 49
 Stages of Recrystallization
33% CW Brass 3 s 580 oC Initial Recryst.

Chapter 7 -
 Stages of Recrystallization
4 s 580 oC Recryst. Cont’d 8 s 580 oC Recryst. Complete

Chapter 7 - 51
 Stages of Recrystallization
15 min 580 oC Grain Growth 10 min 700 oC Grain Growth

Chapter 7 - 52
 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
55 000 psi (380 MPa) and a ductility of at least 15
%EL are desired. Further more, the final diameter
mustt be
b 0.30
0 30 in
i (7.6
(7 6 mm).
) Explain
E l i how
h thi may be
this b
accomplished.

Chapter 7 -
 Cold Work Calculations Solution
If we directly draw to the final diameter
what happens?
Brass
Cold
Work

Do = 0.40 in Df = 0.30 in

 Ao  Af  A 
%CW    x 100  1  f  x 100
 Ao   Ao 
 Df2 4    0.30  2 
 1   x 100  1     x 100  43.8%
 Do 4 
2   0.40  
 
Chapter 7 -
 Coldwork Calc Solution: Cont.

420 540

• For %CW = 43.8%

– y = 420 MPa
– TS = 540 MPa > 380 MPa
– %EL = 6 < 15
• This doesn’t satisfy criteria…… what can we do?
Chapter 7 -
 Coldwork Calc Solution: Cont.

380 15

12 27

For TS > 380 MPa %CW

> 12 %CW
For %EL < 15 %CW
< 27 %CW

%CW = 12
 Our working range is limited to %CW 12--27
Chapter 7 - 56
 Coldwork Calc Soln
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)?
– mustt be
b calculated
l l t d as follows:
f ll
 Df 2 2  Df 2
2
%CW
%CW  1  
2 
x 100  1  2

 D02  D02
100

Df 2  %CW 
0.5 Df 2
 1    D02   %CW  0.5
D02  100  1  
 100 
0 .5
 20 
I t
Intermediate t = Df 1  D02  0.30 1 
di t diameter
di   0.335 m
 100 
Chapter 7 - 57
 Cold Work Calculations: Solution
Summary:
1. Cold work D01= 0.40 in  Df1 = 0.335 m
 2 
 
%CW1  1 
0.335 
x 100  30
  0.4 
 
 

2. Anneal above D02 = Df1

3. Cold work D02= 0.335 in  Df 2 =0.30
0.30 m
  0 .3  2   y  340 MPa
%CW2  1   
  0.335  
 x 100  20 
  TS  400 MPa
%EL  24
Therefore, meets all requirements

Chapter 7 -
 Rate of Recrystallization
E 50%
logR  logt  logR0  start
kT
B 1
logt  C  TR finish
T
note : R  1 / t

• Hot work  above TR log t

• Cold work  below TR

• Smaller
S ll grainsi
– stronger at low temperature
– weaker at high temperature

Chapter 7 - 59
 Summary

• Dislocations are observed primarily in metals

andd alloys.
ll
• 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
H i (annealing)
( li ) can reduce
d di
dislocation
l i d density
i
and increase grain size. This decreases the strength.

Chapter 7 - 60
 Homework

Problems: 7.9, 7.12, 7.13, 7.15, 7.17, 7.19

7.27, 7.29, 7.31, 7.32, 7.36, 7.41,
7.D3, 7.D4.

Due date:

Chapter 7 - 61
Advanced Topics in Dislocation
Phenomena

Chapter 7 - 62
Slip in a Perfect Lattice

First hint as to why there

sshould
ould be crystalline
c ystalli e defects
involved!!!

Chapter 7 - 63
Slip in a Perfect Lattice

Chapter 7 - 64
Slip in a Perfect Lattice

Chapter 7 - 65
Atomic movements near dislocation in
slip (plastic deformation)

Chapter 7 - 66
Why ceramics do not exhibit plastic deformation
at ambient Temperatures?

Chapter 7 - 67
Why ceramics do not exhibit plastic deformation
at ambient Temperatures?

Chapter 7 - 68
Why ceramics do not exhibit plastic deformation
at ambient Temperatures?

Chapter 7 - 69
Stages of Dislocation slip in FCC Metals

Chapter 7 - 70
Stages of Dislocation slip in FCC Metals

Chapter 7 - 71
Stages of Dislocation slip in FCC Metals

Chapter 7 - 72
 Stresses around an Edge Dislocation

Dislocation line passes through “O”

O and extends out of the plane of
the slide…
Chapter 7 - 73
Stresses around an Edge Dislocation

Chapter 7 - 74
Forces on Dislocations

Chapter 7 - 75
Forces on Dislocations

Chapter 7 - 76
Dislocation Sources

Chapter 7 - 77