Transcript Chapter 7: Dislocations & Strengthening Mechanisms ISSUES TO ADDRESS...

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?
Dislocations & Materials Classes
• Metals: Disl. motion easier.
-non-directional bonding
-close-packed directions
for slip.
electron cloud
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ion cores
• Covalent Ceramics
(Si, diamond): Motion hard.
-directional (angular) bonding
• Ionic Ceramics (NaCl):
Motion hard.
-need to avoid ++ and - neighbors.
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Dislocation Motion
Dislocations & plastic deformation
• Cubic & hexagonal metals - plastic deformation is by plastic
shear or slip where one plane of atoms slides over adjacent
plane by defect motion (dislocations).
Adapted from Fig. 7.1,
Callister 7e.
• If dislocations don't move, deformation doesn't occur!
Deformation Mechanisms
Slip System
– Slip plane - plane allowing easiest slippage
• Wide interplanar spacings - highest planar densities
– Slip direction - direction of movement - Highest
linear densities
Adapted from Fig.
7.6, Callister 7e.
– 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
Slip Systems in Cubic Metals:
Stress and Dislocation Motion
• Crystals slip due to a resolved shear stress, tR.
• Applied tension can produce such a stress.
Applied tensile
stress: s = F/A
A
F
Resolved shear
stress: tR =Fs /A s
slip plane
tR = FS /AS
tR
normal, ns
AS
FS
F
Relation between
s and tR
tR
tR  s cos  cos 
Fcos 
F

FS
A/cos 
nS 
A
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
Critical Resolved Shear Stress
• Condition for dislocation motion:
tR  tCRSS
• Crystal orientation can make
it easy or hard to move dislocation
tR  s cos  cos 
s
tR = 0
 =90°
s
typically
10-4 GPa to 10-2 GPa
s
tR = s/2
 =45°
 =45°
tR = 0
 =90°
Generally:
Resolved t (shear stress) is
maximum at  =  = 45º
And
tCRSS = sy/2 for dislocations to
move (in single crystals)
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)
• In General for cubic xtals, angles between directions are
given by:

  Cos 1 


u
2
1
u1u2  v1v2  w1w2
 
 v12  w12  u22  v22  w22





• 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 tR equation to determine
if slip is expected
Slip Motion in Polycrystals
• Stronger since grain boundaries
pin deformations
s
• Slip planes & directions
(, ) change from one
crystal to another.
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].)
• tR will vary from one
crystal to another.
• The crystal with the
largest tR yields first.
• Other (less favorably
oriented) crystals
yield (slip) later.
300 mm
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
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. 7.14 is from A Textbook of Materials
Technology, by Van Vlack, Pearson Education,
Inc., Upper Saddle River, NJ.)
syield  so  k y d 1 / 2
Hall-Petch equation:
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.
Strategies for Strengthening:
2: Solid Solutions
 Impurity atoms distort the lattice & generate stress.
 Stress can produce a barrier to dislocation motion.
• Smaller substitutional
impurity
• Larger substitutional
impurity
A
C
B
Impurity generates local stress at A
and B that opposes dislocation
motion to the right.
D
Impurity generates local stress at C
and D that opposes dislocation
motion to the right.
Stress Concentration at
Dislocations
Adapted from Fig. 7.4,
Callister 7e.
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.
Strengthening by alloying
• Large impurities concentrate at dislocations on
“Tensile Stress” side – pinning dislocation
Adapted from Fig.
7.18, Callister 7e.
Ex: Solid Solution Strengthening in Copper
400
300
200
0 10 20 30 40 50
Yield strength (MPa)
Tensile strength (MPa)
• Tensile strength & yield strength increase with wt% Ni.
wt.% Ni, (Concentration C)
1/ 2
s
~
C
• Empirical relation:
y
• Alloying increases sy and TS.
180
120
60
0 10 20 30 40 50
wt.%Ni, (Concentration C)
Adapted from Fig.
7.16 (a) and (b),
Callister 7e.
Strategies for Strengthening: 3. Precipitation Strengthening
• Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC
in Iron or Aluminum).
precipitate
Large shear stress needed to move
dislocation toward precipitate and
shear it.
Side View
Top View
Unslipped part of slip plane
S
Slipped part of slip plane
• Result:
1
sy ~
S
Dislocation “advances” but
precipitates act as “pinning” sites
with spacing S. which “multiplies”
Dislocation density
Application: Precipitation Strengthening
• Internal wing structure on Boeing 767
Adapted from chapteropening 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.
Adapted from Fig.
11.26, Callister 7e.
(Fig. 11.26 is courtesy
of G.H. Narayanan
and A.G. Miller,
Boeing Commercial
Airplane Company.)
1.5mm
Strategies for Strengthening: 4. Cold Work (%CW)
• Room temperature deformation.
• Common forming operations change the cross
sectional area:
-Forging
roll
die
A o blank
-Drawing
die
Ao
-Rolling
force
Ad
Ao
Adapted from Fig.
11.8, Callister 7e.
roll
-Extrusion
force
Ad
Ad
Ao
tensile
force
force
die
container
ram
billet
container
Ao  Ad
%CW 
x 100
Ao
die holder
extrusion
die
Ad
During Cold Work
• Ti alloy after cold working:
• Dislocations entangle and
multiply
• Thus, Dislocation motion
becomes more difficult.
0.9 mm
Adapted from Fig.
4.6, Callister 7e.
(Fig. 4.6 is courtesy
of M.R. Plichta,
Michigan
Technological
University.)
Result of Cold Work
Dislocation density = total dislocation length
unit volume
– Carefully grown single crystal
 ca. 103 mm-2
– Deforming sample increases density
 109-1010 mm-2
– Heat treatment reduces density
 105-106 mm-2
s
• Yield stress increases sy1
s
y0
as rd increases:
large hardening
small hardening
e
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.
Cold Work Analysis
• What is the tensile strength &
ductility after cold working?
Copper
Cold
Work
D o =15.2mm
D d =12.2mm
2
2
ro  rd
%CW 
x 100  35.6%
2
ro
Cold Work Analysis
• What is the tensile strength &
ductility after cold working to 35.6%?
yield strength (MPa)
tensile strength (MPa)
700
800
500
600
300
100
0
Cu
20
40
% Cold Work
60
YS = 300 MPa
60
40
20
400 340MPa
200
0
ductility (%EL)
20
Cu
40
60
% Cold Work
TS = 340MPa
Cu
7%
00
20
40
60
% Cold Work
%EL = 7%
Adapted from Fig. 7.19, Callister 7e. (Fig. 7.19 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.)
s- e Behavior vs. Temperature
800
Stress (MPa)
• Results for
polycrystalline iron:
Adapted from Fig. 6.14,
Callister 7e.
-200C
600
-100C
400
25C
200
0
0
0.1
0.2
0.3
0.4
Strain
• sy and TS decrease with increasing test temperature.
• %EL increases with increasing test temperature.
3. disl. glides past obstacle
• Why? Vacancies
2. vacancies
help dislocations
replace
move past obstacles. atoms on the
obstacle
disl. half
plane
1. disl. trapped
by obstacle
0.5
Effect of Heating After %CW
• 1 hour treatment at Tanneal...
decreases TS and increases %EL.
• Effects of cold work are reversed!
100 200 300 400 500 600 700
600
60
tensile strength
50
500
40
400
30
ductility
300
20
ductility (%EL)
tensile strength (MPa)
annealing temperature (ºC)
• 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.)
Recovery
Annihilation reduces dislocation density.
• Scenario 1
Results from
diffusion
• Scenario 2
extra half-plane
of atoms
atoms
diffuse
to regions
of tension
extra half-plane
of atoms
3. “Climbed” disl. can now
move on new slip plane
2. grey atoms leave by
vacancy diffusion
allowing disl. to “climb”
1. dislocation blocked;
can’t move to the right
Dislocations
annihilate
and form
a perfect
atomic
plane.
tR
4. opposite dislocations
meet and annihilate
Obstacle dislocation
Recrystallization
• New grains are formed that:
-- have a low dislocation density
-- are small
-- consume cold-worked grains.
0.6 mm
0.6 mm
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.
Further Recrystallization
• All cold-worked grains are consumed.
0.6 mm
0.6 mm
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
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
Adapted from
Fig. 7.21 (d),(e),
Callister 7e.
(Fig. 7.21 (d),(e)
are courtesy of
J.E. Burke,
General Electric
Company.)
After 8 s,
580ºC
After 15 min,
580ºC
coefficient dependent on
material & Temp.
• Empirical Relation:
exponent typ. ~ 2
d  d  Kt
n
grain dia. At time t.
n
o
elapsed time
This is: Ostwald Ripening
º
TR = recrystallization
temperature
TR
Adapted from Fig.
7.22, Callister 7e.
º
Recrystallization Temperature, TR
TR = recrystallization temperature = point
of highest rate of property change
1. TR  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
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.
Coldwork Calculations Solution
If we directly draw to the final diameter what
happens?
Brass
Cold
Work
Do = 0.40 in
 Ao  Af
%CW  
 Ao
Df = 0.30 in


Af 
 x 100  1 
 x 100

 Ao 
  0.30  2 
 Df2 4 
 x 100  43.8%
 x 100  1  
 1 

2
  0.40  
 Do 4 


Coldwork Calc Solution: Cont.
420
540
6
Adapted from Fig.
7.19, Callister 7e.
• For %CW = 43.8%
– sy = 420 MPa
– TS = 540 MPa > 380 MPa
– %EL = 6
< 15
• This doesn’t satisfy criteria…… what can we do?
Coldwork Calc Solution: Cont.
15
380
27
12
For TS > 380 MPa
> 12 %CW
For %EL > 15
< 27 %CW
Adapted from Fig.
7.19, Callister 7e.
 our working range is limited to %CW = 12 – 27%
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:
 D f 22 
%CW  1 
x 100
2 


Ds 2 

 %CW 
 1 

Ds 2 
100 
Df 2
Intermediate diameter =
0.5

 1
Ds 2 
D f 22
Ds 2 2

%CW
100
Df 2
 %CW 
1 

100 

20 

D f 1  Ds 2  0.30 1 

 100 
0.5
0.5
 0.335 in

Coldwork Calculations Solution
Summary:
1. Cold work
2.
3.
D01= 0.40 in  Df1 = 0.335 in
Anneal above Ds2 = Df1
Cold work Ds2= 0.335 in  Df 2 =0.30 in

2 


0.335 
%CW1  1 
x 100  30
  0.4 




  0.3 
%CW2  1  
  0.335 

2

 x 100  20


s y  340 MPa

TS  400 MPa
%EL  24
Fig 7.19
Therefore, meets all requirements
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.