Lecture 6
1. Strengthening and
recrystallization of
plastically deformed
metals.
2. Material selection
according to the
mechanical properties
Why study strengthening
mechanisms?
 We
can tailor the mechanical properties
of an engineering material by choosing
between strength and toughness
Plastic Deformation
 Plastic
deformation is permanent, and strength
and hardness are measures of a material’s
resistance to this deformation.
 On a microscopic scale, plastic deformation
corresponds to the net movement of large
numbers of atoms in response to an applied
stress.
 In crystalline solids, plastic deformation most
often involves the motion of dislocations
 The ability of a metal to plastically deform
depends on the ability of dislocations to move
Dislocation Motion
 Cubic
4
& 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!
Dislocation Motion
5
 The
process by which plastic deformation is
produced by dislocation motion is termed slip
 Dislocation moves along slip plane in slip
direction perpendicular to dislocation line
Edge dislocation
Screw dislocation
Strengthening of Metals
 Good
industrial alloys -> high strengths yet
some ductility and toughness
 Since hardness and strength are related to the
ease with which plastic deformation can be
made to occur, by reducing the mobility of
dislocations, the mechanical strength may be
enhanced
 In contrast, the more unconstrained the
dislocation motion, the greater is the facility
with which a metal may deform, and the
softer and weaker it becomes
 Restricting or hindering dislocation motion
renders a material harder and stronger
Strengthening Methods
 grain
size reduction,
 solid-solution alloying,
 precipitation hardening/strengthening
 strain hardening/strengthening
8
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:
 yield   o  k y d
1 / 2
d is the average grain diameter. σo and ky
are constants for a particular material
Dependence of Yield
Strength on Grain Size
The influence of grain size on the yield
strength of a 70 Cu–30 Zn brass alloy
10
Strategies for Strengthening:
2: Solid Solution Strengthening
• 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
11
Stress Concentration at
Dislocations
SSS - Impurity Atoms
SSS – Effects of Impurity
Atoms
 The
resistance to slip is greater when impurity
atoms are present because the overall lattice
strain must increase if a dislocation is moved
away from them.
 The same lattice strain interactions will exist
between impurity atoms and dislocations that
are in motion during plastic deformation.
 Thus, a greater applied stress is necessary to
first initiate and then continue plastic
deformation for solid-solution alloys, as
opposed to pure metals
SSS – Strength and Ductility
Variation with nickel content of (a) tensile
strength, (b) yield strength, and (c) ductility
(%EL) for copper–nickel alloys, showing
strengthening.
Strategies for Strengthening:
3: Precipitation Strengthening
 Precipitation
strengthening, also called age
hardening, is a heat treatment technique used to
increase the yield strength of malleable materials.
 It relies on changes in solid solubility with
temperature to produce fine particles of an
impurity phase, which impede the movement of
dislocations, or defects in a crystal's lattice.
 Precipitation in solids can produce many different
sizes of particles, which have different properties.
 Alloys must be kept at elevated temperature for
hours to allow precipitation to take place. This
time delay is called aging
16
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:
y~
1
S
Dislocation
“advances” but
precipitates act as
“pinning” sites with
spacing S.
17
Application:
Precipitation Strengthening
• Internal wing structure on Boeing 767
• Aluminum is strengthened with precipitates formed
by alloying.
1.5mm
18
Strategies for Strengthening:
4: Strain-Hardening
• Room temperature deformation.
• Common forming operations change the cross
sectional area:
-Forging
force
die
A o blank
-Rolling
roll
Ao
Ad
Ad
roll
-Drawing
die
Ao
force
Ad
-Extrusion
Ao
tensile
force
force
die
container
ram
billet
container
Ao  Ad
% CW 
x 100
Ao
die holder
extrusion
die
Ad
19
Dislocations During Cold Work
• Ti alloy after cold working:
• Dislocations entangle
with one another
during cold work.
• Dislocation motion
becomes more difficult.
0.9 mm
Result of Cold Work
20
total dislocation length
Dislocation density =
unit volume
 Carefully
grown single crystal
 103 mm-2
 Deforming sample increases density
 109-1010 mm-2
 Heat treatment reduces density
 105-106 mm-2
• Yield stress increases
as dislocation density
increases:

y1
y0
large hardening
small hardening
e
Impact of Cold Work
21
As cold work is increased
• Yield strength (y) increases.
• Tensile strength (TS) increases.
• Ductility (%EL or %AR) decreases.
22
Cold Work Analysis
• What is the tensile strength &
ductility after cold working?
2
Copper
Cold
Work
2
 ro   rd
% CW 
x 100  35 . 6 %
2
 ro
yield strength (MPa)
tensile strength (MPa)
700
800
500
600
300
300MPa
100
0
20
Cu
40
% Cold Work
y = 300MPa
60
Do =15.2mm
Dd =12.2mm
ductility (%EL)
60
40
Cu
20
400 340MPa
200
0
Cu
7%
20
40
60
% Cold Work
TS = 340MPa
00
20
40
60
% Cold Work
%EL = 7%
-e Behavior vs. Temperature
23
800
Stress (MPa)
• Results for
polycrystalline iron:
-200C
600
-100C
400
25C
200
0
0
0.1
0.2
0.3
0.4
Strain
• y 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
24
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...
25
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
 Even
after recovery is complete, the grains
are still in a relatively high strain energy
state
 Recrystallization is the formation of a new
set of strain-free and equiaxed grains (i.e.,
having approximately equal dimensions in
all directions) that have low dislocation
densities and are characteristic of the
precold-worked condition.
 The new grains form as very small nuclei
and grow until they completely consume
the parent material, processes that involve
short-range diffusion
Recrystallization
27
• New grains are formed that:
-- have a small dislocation density
-- are small
-- consume cold-worked grains.
0.6 mm
33% cold
worked
brass
0.6 mm
New crystals
nucleate after
3 sec. at 580C.
Further Recrystallization
28
• All cold-worked
grains are consumed.
0.6 mm
After 4
seconds
0.6 mm
After 8
seconds
Recrystallization
 During
recrystallization, the mechanical properties
that were changed as a result of cold working are
restored to their precold-worked values; that is,
the metal becomes softer, weaker, yet more
ductile
 Recrystallization is a process the extent of which
depends on both time and temperature. The
degree (or fraction) of recrystallization increases
with time
 For pure metals, the recrystallization temperature
is normally 0.3Tm where Tm is the absolute melting
temperature
Grain Growth
 After
recrystallization is complete, the
strain-free grains will continue to grow if
the metal specimen is left at the elevated
temperature; this phenomenon is called
grain growth
 Grain growth does not need to be
preceded by recovery and
recrystallization; it may occur in all
polycrystalline materials, metals and
ceramics alike
Grain Growth
31
• At longer times, larger grains consume smaller ones.
• Why? Grain boundary area (and therefore energy)
is reduced.
0.6 mm
After 8 s,
580ºC
0.6 mm
After 15 min,
580ºC
• Empirical Relation:
exponent typical. ~ 2
grain diam.
n
n
d

d
 Kt
at time t.
o
After 10 min,
700ºC
coefficient dependent
on material and T.
elapsed time
Ostwald Ripening
º
32
TR = recrystallization
temperature
TR
º
Time and Temperature
Dependent Grain Growth
Recrystallization
Temperature, TR
34
TR = recrystallization temperature = point of
highest rate of property change
1. Tm => TR  0.3-0.6 Tm (K)
2. Due to diffusion  annealing time TR = f(t)
shorter annealing time => higher TR
3. Pure metals lower TR due to dislocation
movements

Easier to move in pure metals => lower TR
35
Coldwork Calculations
A cylindrical rod of brass originally 0.40in
(10.2mm) in diameter is to be cold worked
by drawing. The circular cross section will be
maintained during deformation. A coldworked tensile strength in excess of
55,000psi (380MPa) and a ductility of at least
15%EL are desired. Furthermore, the final
diameter must be 0.30in (7.6mm). Explain
how this may be accomplished.
36
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


A o 


2


Df 4 
1 

  1 
x
100

2



D
4
o



2
 0 . 30  

 x 100  43 . 8 %
 0 . 40  
37
Coldwork Calc Solution:
Cont.
420
540
6
 For %CW = 43.8%
– y = 420 MPa
– TS = 540 MPa
> 380 MPa
– %EL = 6 < 15
•
This doesn’t satisfy criteria…… what can we do?
38
Coldwork Calc Solution:
Cont.
15
380
27
12
For TS > 380 MPa
> 12 %CW
For %EL < 15
< 27 %CW
 our working range is limited to %CW = 12-27
39
Coldwork Calc 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)?
 must be calculated as follows:
2

Df 2
% CW   1 
2

D
02

Df 2
D 02
% CW 

 1 

100 

Intermediate diameter =

 x 100


0 .5

D f 1  D 02

1
Df 2
D 02
D 02 
2
2

% CW
100
Df 2
0 .5
% CW 

1 

100 0. 5

20 

 0 . 30  1 

100 

 0 . 335 in
40
Coldwork Calculations
Solution
Summary:
1. Cold work
D01= 0.40 in  Df1 = 0.335 in
2

0
.
335

 

% CW 1  1  
 x 100  30 %

 0 . 4  

2.
3.
Anneal to remove all CW effects; D02 = Df1
Cold work D02= 0.335 in  Df 2 =0.30 in
2

0
.
3

 

% CW 2  1  
 x 100  20 %

 0 . 335  


Therefore, meets all requirements

y
 340 MPa
TS  400 MPa
% EL  24
Material Selection
According to the Mechanical
Properties
Material Selection – The Basics
 Getting
the optimum balance of performance,
quality, and cost requires a careful combination
of material and part design
 The ideal product is one that will just meet all
requirements.
 Anything better will usually incur added cost
through higher-grade materials, enhanced
processing, or improved properties that may not
be necessary.
 Anything worse will likely cause product failure,
dissatisfied customers, and the possibility of
unemployment
Material Selection – The Basics
 The
interdependence between materials and
their processing must also be recognized.
 New processes frequently accompany new
materials, and their implementation can often
cut production costs and improve product
quality.
 A change in material may well require a
change in the manufacturing process
 Improper processing of a well-chosen material
can definitely result in a defective product.
Materials and
Manufacturing
• An engineering material may possess different properties
depending upon its structure.
• Processing of that material can alter the structure, which in turn
will alter the properties.
• Altered properties certainly alter performance.
PROCEDURES FOR MATERIAL
SELECTION
Every engineered item goes through a sequence
of activities that includes:
 design
 material selection
 process selection
 production
 evaluation
 possible redesign or modification
Requirements for Material
Selection
(1) shape or geometry considerations,
(2) property requirements,
(3) manufacturing concerns
1. GEOMETRIC CONSIDERATIONS
 What
is the relative size of the component?
 How complex is its shape?
 What are the surface-finish requirements?
Must all surfaces be finished?
 Could a minor change in part geometry
increase the ease of manufacture or improve
the performance (fracture resistance, fatigue
resistance, etc.) of the part?
2. Mechanical Properties
 How
much static strength is required?
 If the part is accidentally overloaded, is it
permissible to have a sudden brittle fracture, or
is plastic deformation and distortion a desirable
precursor to failure?
 How much can the material bend, stretch, twist,
or compress under load and still function
properly?
 Are any impact loadings anticipated? If so, of
what type, magnitude, and velocity?
2. Mechanical Properties
 Can
you envision vibrations or cyclic
loadings? If so, of what type, magnitude,
and frequency?
 Is wear resistance desired? Where? How
much? How deep?
 Will all of the above requirements be
needed over the entire range of
operating temperature? If not, which
properties are needed at the lowest
extreme? At the highest extreme?
Environmental Considerations
 What
are the lowest, highest, and normal
temperatures the product will see? Will
temperature changes be cyclic? How fast will
temperature changes occur?
 What is the most severe environment that is
anticipated as far as corrosion or deterioration
of material properties is concerned?
 What is the desired service lifetime for the
product?
3. Manufacturing Concerns
 How
many of the components are to be
produced? At what rate?
 What is the desired level of quality compared to
similar products on the market?
 Has the design addressed the requirements that
will facilitate ease of manufacture?
Ashby Material Selection
Charts
E MCH 213D