Chapter 7:
description
Transcript of Chapter 7:
Chapter 7:Dislocations & Strengthening Mechanisms
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WHY STUDY Dislocations and Strengthening Mechanisms?Understand underlying mechanisms
of the techniques used to strengthen and harden metals and their alloys.
Design and 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 and more ductile and produces changes in its microstructure.
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1. Describe edge and 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 and 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 and explain phenomenon of strain hardening in terms of dislocations and strain field interactions.
7. Describe recrystallization in terms of both alteration of microstructure and mechanical characteristics of the material.
8. Describe phenomenon of grain growth from both macroscopic and atomic perspectives.
Learning Objectives
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7.2 BASIC CONCEPTSDislocation 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 the 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. As a result, 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 the half-plane, the lattice atoms sustain an imposed tensile strain.
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7.3 Characteristics of Dislocations
Shear strains also exist in the vicinity of the 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 Dislocationsstrain fields surrounding
Consider two edge dislocations that have the same sign and the identical slip plane (Figure 7.5a).
Compressive and tensile strain fields for both lie on the 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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7.3 Characteristics of Dislocations
Two dislocations of opposite sign and having same slip plane will be attracted to one another (Figure 7.5b), the two extra half-planes of atoms will align and become a complete plane.
These strain fields and associated forces are important in the strengthening mechanisms for metals.
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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 normal to slip plane & stress direction
slip planenormal, ns
Resolved shear stress: R =Fs/As
slipdire
ction
AS
R
R
FS
slipdire
ction
Relation between and R
R =FS /AS
Fcos A/cos
F
FS
nS
AS
A
Applied tensile stress: = F/A
slipdire
ction
FA
F
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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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16Critical 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 and angles for Slip in Crystals
For cubic xtals, angles between directions are given by:
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 R equation to determine if slip is expected
1 1 2 1 2 1 2
2 2 2 2 2 21 1 1 2 2 2
u u v v w wCosu v w u v w
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EXAMPLE PROBLEM 7.1Resolved Shear Stress and Stress-to-Initiate-Yielding ComputationsConsider a single crystal of BCC iron oriented such that a tensile stress is applied along a [010] direction.Compute the resolved shear stress along a (110) plane and in a direction when a tensile stress of 52 MPa is applied.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.
The crystal with the largest R yields first.
Other (less favorably oriented) crystals yield (slip) later.
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7.6 Plastic Deformation ofPolycrystalline Materials
300 m
7.6 Plastic Deformation of Polycrystalline Materials
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:1. grain size reduction2. solid-solution alloying3. 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
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.
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7.8 Strengthening by Grain Size ReductionGrain Size Reduction Techniques:
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~
Tens
ile s
treng
th (M
Pa)
wt.% Ni, (Concentration C)
200
300
400
0 10 20 30 40 50
Yie
ld s
treng
th (M
Pa)
wt.%Ni, (Concentration C)
60
120
180
0 10 20 30 40 50
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287.9 Solid-solution Strengthening
7.10 Strain Hardening Room temperature deformation. Common forming operations change the cross
sectional area:-Forging
Ao Ad
force
dieblank
force-Drawing
tensile force
AoAddie
die
-Extrusion
ram billet
container
containerforce die holder
die
Ao
Adextrusion
100 x %o
doA
AACW
-Rolling
roll
AoAd
roll
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Ti alloy after cold working:Dislocations entangle and multiplyThus, Dislocation motion becomes more difficult.
0.9 m
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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 lengthunit volume
large hardeningsmall hardening
y0 y1
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7.10 Strain Hardening
As cold work is increased Yield strength (y) increases.Tensile strength (TS) increases.Ductility (%EL or %AR) decreases.
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7.10 Strain Hardening
• What is the tensile strength & ductility after cold working?
%6.35100 x % 2
22
o
dor
rrCW
Cold Work Analysis
D o =15.2mm
Cold Work
Dd =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
200 40 60
yield strength (MPa)
% Cold Work
tensile strength (MPa)
200Cu
0
400
600
800
20 40 60
340MPa
TS = 340MPa
ductility (%EL)
% Cold Work
20
40
60
20 40 6000
Cu
7%
%EL = 7%YS = 300 MPa
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Cold Work Analysis
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.
- Behavior vs. Temperature
2. vacancies replace atoms on the disl. half plane
3. disl. glides past obstacle
-200C
-100C
25C
800
600
400
200
0Strain
Stre
ss (M
Pa)
0 0.1 0.2 0.3 0.4 0.5
1. disl. trapped by obstacle
obstacle
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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,and Grain Growth
tens
ile s
treng
th (M
Pa)
duct
ility
(%E
L)tensile strength
ductility
Recovery
Recrystallization
Grain Growth
600
300
400
500
60
50
40
30
20
annealing temperature (ºC)200100 300 400 500 600 700
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Annihilation reduces dislocation density.7.11 Recovery
• Scenario 1Results 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
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New grains are formed that:have a low dislocation densityare smallconsume cold-worked grains.
33% coldworkedbrass
New crystalsnucleate after3 sec. at 580C.
0.6 mm 0.6 mm
7.12 Recrystallization04/22/23 07:59
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• All cold-worked grains are consumed.
After 4seconds
After 8seconds
0.6 mm0.6 mm
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7.12 Recrystallization
7.12 Recrystallization
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TR
º
º
TR = recrystallization temperature
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DESIGN EXAMPLE 7.1Description of Diameter Reduction ProcedureA 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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At longer times, larger grains consume smaller ones.
Why? 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 no
n
coefficient dependent on material & Temp.
grain dia. At time t.elapsed time
exponent typ. ~ 2
This is: Ostwald Ripening
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Recrystallization Temperature, TR
TR = recrystallization temperature = point of highest rate of property change1. 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
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Coldwork CalculationsA 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 4003001100 x
441
100 1100 x %
2
2
2
...
DD
xAA
AAACW
o
f
o
f
o
fo
Do = 0.40 in
BrassCold Work
Df = 0.30 in
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Coldwork Calc Solution: Cont.
For %CW = 43.8%
540420
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 RecrystallizationCold 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 22 2
2 2
%% 1 100 1100
f f
s s
D D CWCW xD D
0.5
2
2
%1100
f
s
D CWD
22 0.5%1
100
fs
DD
CW
0.5
1 2200.30 1 0.335 in
100f sD D
Intermediate diameter =
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Coldwork Calculations SolutionSummary:1. Cold work D01= 0.40 in Df1 = 0.335 in2. 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
ELTS
y
%CW1 10.335
0.4
2
x 100 30
Fig 7.19
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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 size2. solid solution strengthening3. precipitate strengthening4. 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 Monday 21/11/2011
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