Chapter 7: Dislocations & Strengthening Mechanismssite.iugaza.edu.ps/mhaiba/files/2010/02/CH7... ·...
Transcript of Chapter 7: Dislocations & Strengthening Mechanismssite.iugaza.edu.ps/mhaiba/files/2010/02/CH7... ·...
Chapter 7 Dislocations & Strengthening Mechanisms
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5:41 AM
Dr. Mohammad Suliman Abuhaiba, PE 1
WHY STUDY Dislocations and Strengthening Mechanisms?
Understand underlying mechanisms
of the techniques used to
strengthen & harden metals and
their alloys
Design & tailor mechanical
properties of materials
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WHY STUDY Dislocations and Strengthening Mechanisms?
Strengthening mechanisms will be
applied to the development of
mechanical properties for steel alloys
Why heat treating a deformed metal
alloy makes it softer & more ductile
and produces changes in its
microstructure
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1. Describe edge & screw dislocation motion
2. Describe how plastic deformation occurs by
motion of edge & screw dislocations in response
to applied shear stresses
3. Define slip system
4. Describe how grain structure of a polycrystalline
metal is altered when it is plastically deformed.
5. Explain how grain boundaries impede dislocation
motion & why a metal having small grains is
stronger than one having large grains.
Learning Objectives
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5. Describe & explain solid-solution strengthening
for substitutional impurity atoms in terms of lattice
strain interactions with dislocations
6. Describe & explain phenomenon of strain
hardening in terms of dislocations and strain field
interactions.
7. Describe recrystallization in terms of both
alteration of microstructure & mechanical
characteristics of the material
8. Describe phenomenon of grain growth from both
macroscopic and atomic perspectives.
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Learning Objectives
7.2 BASIC CONCEPTS
Dislocation Motion Cubic & hexagonal metals - plastic deformation is
by plastic shear or slip where one plane of atoms
slides over adjacent plane by dislocations motion.
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7.3 Characteristics of Dislocations
When metals are plastically
deformed:
Around 5% of deformation energy is
retained internally
remainder is dissipated as heat
The major portion of stored
energy is strain energy associated
with dislocations
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7.3 Characteristics of Dislocations
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Figure 7.4: some atomic lattice distortion exists
around dislocation line because of presence of
extra half-plane of atoms.
7.3 Characteristics of Dislocations
Atoms immediately above and
adjacent to dislocation line are
squeezed together.
These atoms may be thought of as
experiencing a compressive strain relative
to atoms positioned in the perfect crystal
and far removed from the dislocation
Directly below half-plane, lattice
atoms sustain an imposed tensile strain
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7.3 Characteristics of Dislocations
Shear strains also exist in the vicinity of
edge dislocation
For a screw dislocation, lattice strains are
pure shear only
These lattice distortions may be
considered to be strain fields that radiate
from the dislocation line
The strains extend into the surrounding
atoms, and their magnitude decreases
with radial distance from the dislocation.
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7.3 Characteristics of Dislocations Strain fields surrounding
Figure 7.5 (a): Two edge dislocations of same sign
& lying on same slip plane exert a repulsive force
on each other
Compressive & tensile strain fields for both lie on
same side of the slip plane; there exists between
these two isolated dislocations a mutual repulsive
force that tends to move them apart
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Figure 7.5 (b) Edge dislocations of opposite
sign & lying on same slip plane exert an
attractive force on each other. Upon
meeting, they annihilate each other and
leave a region of perfect crystal
7.3 Characteristics of Dislocations strain fields surrounding
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7.4 Slip Systems Slip plane - plane allowing easiest slippage
Wide inter-planar spacing - highest planar densities
Slip direction - direction of movement - Highest
linear densities
FCC Slip occurs on {111} planes in <110> directions
=> total of 12 slip systems in FCC
in BCC & HCP other slip systems occur
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Table 7.1: Slip Systems for FCC,
BCC, and HCP Metals
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7.5 Slip in Single Crystals
coscosR
Crystals slip due to a
resolved shear stress, R.
Applied tension can produce
such a stress.
= angle between stress
direction & Slip direction
= angle between stress
direction & normal to slip
plane
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Condition for dislocation motion: CRSS R
Crystal orientation can make it
easy or hard to move dislocation
10-4 GPa to 10-2 GPa
typically
coscosR
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Critical Resolved Shear Stress
R = 0
=90°
R = /2 =45° =45°
R = 0
=90°
7.5 Slip in Single Crystals
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Resolved (shear stress) is maximum at = = 45º
And
CRSS = y/2 for dislocations to move (in single crystals)
Determining & angles for Slip in
Crystals
For cubic crytals, angles between directions are given by:
1 1 2 1 2 1 2
2 2 2 2 2 2
1 1 1 2 2 2
u u v v w wCos
u v w u v w
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EXAMPLE PROBLEM 7.1 Resolved Shear Stress and Stress-to-Initiate-Yielding
Computations
Consider a single crystal of BCC iron oriented such
that a tensile stress is applied along a [010] direction.
a. Compute the resolved shear stress along a (110)
plane and in a direction when a tensile stress
of 52 MPa is applied.
b. If slip occurs on a (110) plane and in a [-111]
direction, and the critical resolved shear stress is
30MPa, calculate the magnitude of the applied
tensile stress necessary to initiate yielding.
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Stronger since grain boundaries
pin deformations
Slip planes & directions (, )
change from one crystal to
another
R will vary from one crystal to
another.
Crystal with largest R yields first
Other (less favorably oriented)
crystals yield (slip) later.
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7.6 Plastic Deformation of Polycrystalline Materials
300 mm
7.6 Plastic Deformation of Polycrystalline Materials
Ordinarily ductility is sacrificed when an alloy is strengthened.
Relationship between dislocation motion & mechanical behavior of metals is significance to understanding of strengthening mechanisms
The ability of a metal to plastically deform depends on the ability of dislocations to move.
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7.6 Plastic Deformation of Polycrystalline Materials
Strengthening principle: Restricting or Hindering dislocation motion renders a material harder & stronger.
We will consider strengthening single phase metals by:
1. Grain size reduction
2. Solid-solution alloying
3. Strain hardening
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7.8 Strengthening by Grain Size Reduction
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Grain boundaries are barriers to slip
Barrier "strength" increases with Increasing
angle of mis-orientation.
Smaller grain size: more barriers to slip
Hall-Petch equation:
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21 /
yoyield dk
7.8 Strengthening by Grain
Size Reduction Figure 7.15: Influence of
grain size on yield
strength of a 70 Cu–30 Zn
brass alloy
Increase Rate of solidification from the liquid
phase
Perform Plastic deformation followed by an
appropriate heat treatment
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7.8 Strengthening by Grain Size
Reduction Grain Size Reduction Techniques:
Grain size reduction improves toughness of many
alloys.
Small-angle grain boundaries are not effective in
interfering with the slip process because of small
crystallographic misalignment across the boundary.
Boundaries between two different phases are also
impediments to movements of dislocations.
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7.8 Strengthening by Grain Size Reduction
Impurity atoms distort the lattice & generate stress.
Stress can produce a barrier to dislocation motion.
7.9 Solid-solution Strengthening
• 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
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7.9 Solid-solution Strengthening
small impurities tend to concentrate at
dislocations on the “Compressive stress side”
reduce mobility of dislocation increase
strength
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Large impurities concentrate at dislocations on
“Tensile Stress” side – pinning dislocation
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7.9 Solid-solution Strengthening
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• Tensile strength & yield strength increase with wt% Ni.
• Empirical relation:
• Alloying increases y and TS
21 /
y C~
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7.9 Solid-solution Strengthening
-Forging
A o A d
force
die blank
force
7.10 Strain Hardening Room temperature deformation
Common forming operations change cross
sectional area:
-Drawing
tensile force
A o A d die
die
-Extrusion
ram billet
container
container
force die holder
die
A o
A d extrusion
100 x %o
do
A
AACW
-Rolling
roll
A o A d
roll
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Ti alloy after cold working:
Dislocations entangle &
multiply
Thus, Dislocation motion
becomes more difficult
0.9 mm
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7.10 Strain Hardening
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
e
y0
y1
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7.10 Strain Hardening
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7.10 Strain Hardening
Figure 7.20:
influence of cold
work on stress
strain behavior of a
low-carbon steel
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7.10 Strain Hardening
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7.10 Strain Hardening
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7.10 Strain Hardening
What is the tensile strength & ductility after cold
working?
%6.35100 x %2
22
o
do
r
rrCW
Cold Work Analysis
D o =15.2mm
Cold
Work
D d =12.2mm
Copper
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What is the tensile strength & ductility after cold working to
35.6%?
% Cold Work
100
300
500
700
Cu
20 0 40 60
yield strength (MPa)
% Cold Work
tensile strength (MPa)
200
Cu
0
400
600
800
20 40 60
340MPa
TS = 340MPa
ductility (%EL)
% Cold Work
20
40
60
20 40 60 0 0
Cu
7%
%EL = 7% YS = 300 MPa
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Cold Work Analysis
Results for polycrystalline iron:
y & TS decrease with increasing temperature
%EL increases with increasing temperature
- e Behavior vs. Temperature
-200C
-100C
25C
800
600
400
200
0
Strain
Str
ess (
MP
a)
0 0.1 0.2 0.3 0.4 0.5
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Recovery, Recrystallization,
and Grain Growth
Properties & structures may revert back to
the precold-worked states by appropriate
heat treatment (annealing treatment).
Such restoration results from different
processes that occur at elevated
temperatures:
1. Recovery
2. Recrystallization
3. Grain growth
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1 hour treatment at
Tanneal... decreases TS and
increases %EL.
Effects of cold work
are reversed!
3 Annealing stages
to discuss...
Recovery, Recrystallization, & Grain
Growth
te
nsile
str
en
gth
(M
Pa
)
du
ctilit
y (
%E
L) tensile strength
ductility
600
300
400
500
60
50
40
30
20
annealing temperature (ºC) 200 100 300 400 500 600 700
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7.11 RECOVERY During recovery, some of the stored internal strain
energy is relieved by virtue of dislocation motion,
due to enhanced atomic diffusion at elevated
temperature.
There is some reduction in number of dislocations,
and dislocation configurations are produced
having low strain energies.
Physical properties such as electrical and thermal
conductivities are recovered to their precold-
worked states.
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Even after recovery is complete, the grains are still in a
relatively high strain energy state.
Recrystallization = formation of a new set of strain-
free and equiaxed grains that have low dislocation
densities and are characteristic of the pre-cold-worked
condition.
The driving force to produce this new grain structure is
the difference in internal energy between the strained
and unstrained material.
The new grains form as very small nuclei and grow
until they completely consume the parent material,
processes that involve short-range diffusion.
7.12 Recrystallization
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33% cold
worked
brass
New crystals
nucleate after
3 sec. at 580C.
0.6 mm 0.6 mm
7.12 Recrystallization
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Figure 7.21
All cold-worked grains are consumed.
After 4
seconds
After 8
seconds
0.6 mm 0.6 mm
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7.12 Recrystallization
7.12 Recrystallization
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7.12 Recrystallization
Recrystallization temperature =
temperature at which recrystallization just
reaches completion in 1 h.
Recrystallization temperature for brass alloy of Figure 7.22 is about 450°C.
Typically, between one-third and one-half
of absolute melting temperature of a
metal or alloy
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TR
º
º
TR = recrystallization
temperature
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Figure 7.22: Annealing
temperature influence
(for an annealing time
of 1 h) on tensile
strength & ductility of a
brass alloy.
7.12 Recrystallization
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Figure 7.23
Critical degree of
cold work below
which
recrystallization
cannot be made
to occur (between
2% and 20% cold
work)
Recrystallization proceeds more
rapidly in pure metals than in
alloys.
During recrystallization, grain
boundary motion occurs as the
new grain nuclei form and then
grow.
For pure metals, TRC = 0.4Tm
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7.12 Recrystallization
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7.12 Recrystallization
At longer times, larger
grains consume
smaller ones.
Grain boundary area
(and therefore energy)
is reduced.
Empirical Relation:
After 8 s,
580ºC
After 15 min,
580ºC
0.6 mm 0.6 mm
7.13 Grain Growth
Ktdd n
o
n
coefficient dependent on
material & Temp.
grain dia. At time t.
elapsed time exponent typ. ~ 2
This is: Ostwald Ripening
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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.
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Coldwork Calculations Solution
If we directly draw to the final diameter what
happens?
%843100 x 400
3001100 x
4
41
100 1100 x %
2
2
2
..
.
D
D
xA
A
A
AACW
o
f
o
f
o
fo
D o = 0.40 in
Brass
Cold Work
D f = 0.30 in
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Coldwork Calc Solution: Cont.
For %CW = 43.8%
540 420
– y = 420 MPa
– TS = 540 MPa > 380 MPa
6
– %EL = 6 < 15
• This doesn’t satisfy criteria…… what can we do?
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Coldwork Calc Solution: Cont.
380
12
15
27
For %EL > 15
For TS > 380 MPa > 12 %CW
< 27 %CW
our working range is limited to %CW = 12 – 27%
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This process Needs an Intermediate
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 2
2 2
2 2
2 2
%% 1 100 1
100
f f
s s
D D CWCW x
D D
0.5
2
2
%1
100
f
s
D CW
D
2
2 0.5%
1100
f
s
DD
CW
0.5
1 2
200.30 1 0.335 in
100f sD D
Intermediate diameter =
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Coldwork Calculations Solution Summary:
1. Cold work D01= 0.40 in Df1 = 0.335 in
2. Anneal above Ds2 = Df1
3. Cold work Ds2= 0.335 in Df 2 =0.30 in
Therefore, meets all requirements
20100 3350
301%
2
2
x
.
.CW 24%
MPa 400
MPa 340
EL
TS
y
%CW1 10.335
0.4
2
x 100 30
Fig 7.19
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Design Example 7.1
Description of Diameter Reduction Procedure
A cylindrical rod of non-cold-worked brass having an
initial diameter of 6.4 mm is to be cold worked by
drawing such that the cross-sectional area is
reduced. It is required to have a cold-worked yield
strength of at least 345 MPa and a ductility in excess
of 20 %EL; in addition, a final diameter of 5.1 mm is
necessary. Describe the manner in which this
procedure may be carried out.
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Dislocations are observed primarily in metals and
alloys.
Strength is increased by making dislocation motion
difficult.
Particular ways to increase strength are to:
1. decrease grain size
2. solid solution strengthening
3. precipitate strengthening
4. cold work
Heating (annealing) can reduce dislocation density
and increase grain size. This decreases the strength
.
Summary
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HomeWork Assignment
7, 12, 17, 24, 29, 41, D6
Due Tuesday 26/11/2013
Quiz is on Tuesday 3/12/2013
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