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Plastic deformation and
creep in
crystalline materialsChap. 11
Mechanical Properties of Materials
Stiffness
Strength
ductility
Toughness
Resistance to elastic deformation
Youngs modulus
Resistance to plasticdeformation
Yield stress
Resistance to fracture Energy to fracture
Ability to deform plastically
Strain to fracture
Uniaxial Tensile Test
Gaugelength
specimen
Result of a uniaxial tensile test
Slope = Youngs modulus (Y)
UTSUltimate tensile strength
yYield strength
(Engineering stress)
(engineering strain)
f (strain to fracture)
necking
Area = Toughness
elas
tic
plastic
break
Yield point
STIFFNESS
STRENGTH
DUCTILITY
If there is a smooth transition from elastic to plastic region (no distinct yield point)
then 0.2 % offset proof stress is used
During uniaxial tensile test the length of the specimen is continually increasing and the cross-sectional area is decreasing.
True stress Engineering stress (=F/A0)True strain Engineering strain (=L/L0)
True stress i
T AP
= Ai = instantaneous area
True incremental strain L
dLd T =
0
ln0
LL
LdLL
LT == True strain
Eqn. 11.3
Eqn. 11.4
nTT K =
K Strength coefficientn work hardening exponent
Eqn. 11.5
What happens during plastic deformation?
Externally, permanent shape change begins at y
Internally, what happens?
What happens to crystal structure after plastic deformation?
?Plastic
Deformation
Some Possible answers
Remains the
same
Changes to
another
crystal
structure
Becomes random
or
amorphous
How Do We Decide?
X-ray diffraction
No change in crystal structure!
No change in internal crystal structure but change in external shape!!
How does the microstructure of polycrystalchanges during plastic deformation?
EXPERIMENT 5
Comparison of undeformed Cu and deformed Cu
Slip Lines
Before Deformation After Deformation
Slip lines in the microstructure of
plastically deformed Cu
Callister
Slip
Slip Planes, Slip Directions, Slip Systems
Slip Plane: Crystallographic planes
Slip Direction: Crystallographic direction
Slip System: A combination of a slip plane and a slip direction
Slip Systems in Metallic Crystals
Crystal Slip Slip Slip Plane Direction Systems
FCC {111} 4x3=12(4 planes) (3 per plane)
BCC {110} 6x2=12(6 planes) (2 per plane)
HCP {001} 3x1=3(1 plane) (3 per plane)
Why slip planes are usually close packed planes?
Why slip directions are close-packed directions?
Slip Systems in FCC Crystal
x
y
z(111)
Tensile vs Shear Stress
Plastic deformation takes place by slip
Slip requires shear stress
Then, how does plastic deformation take place during a tensile test?
ND21
: Applied tensile stress
N: Slip plane normal
D: Slip direction
1: angle between and N
2 =angle between and D
Is there any shear stress on the slip plane in the slip direction due to the applied tensile stress?
FND21
F
Area=A = F/ A
FD = F cos 2
Area = As
As = A cos 1
S
DRSS A
F=
1
2
cos
cos
A
F=
21 coscos AF
=
21 coscos =RSS
Resolved Shear stress
FF
F
F
No resolved shear stress on planes parallel or perpendicular to the stress axis
cos 2 = 0 cos 1 = 0
Plastic deformation recap
No change in crystal structure:sliptwinning
Slip takes place on slip systems (plane + direction)Slip planes usually close-packed planesSlip directions usually close-packed direction
Slip requires shear stressIn uniaxial tension there is a shear component of tensile stress on the slip plane in the slip direction:
RESOLVED SHEAR STRESS
Extra ClassesThursdays 10-11
MS 702
21 coscos =RSSRSS
21 coscos y
CRSS
CRITICAL RESOVED SHEAR STRESS
21 coscos yCRSS =
ND21
21 coscos yCRSS =
If we change the direction of stress with respect to the slip plane and the slip direction cos 1 cos 2 will change.
1. CRSS changes.
To maintain the equality which of the following changes takes place?
2. y changes
Schmids Law: CRSS is a material constant.
Anisotropy of Yield Stress
21coscos
crssy =
Yield stress of a single crystal depends upon the direction of application of load
cos 1 cos 2 is called the Schmid factor
RSS
bb2
coscos 1
y
CRSS
Active slip system
21 coscos yCRSS =
aa
2coscos 1
Slip system with highest Schmidfactor is the active slip system
Magnitude of
Critical Resolved Shear StressTheory (Frenkel 1926)
Experiment
bd
CRSS
Shear stress
b/2 b
Potential energy
Fe (BCC)
Cu (FCC)
Zn (HCP)
Theory
(GPa)12
7
5
Experiment
(MPa)15
0.5
0.3
Ratio
Theory/Exp
800
14,000
17,000
Critical Resolved Shear Stress
?
19341934
E. OrowanMichael Polanyi
Geoffrey Ingram Taylor
Solution
Solution
Not a rigid body slip
Part slip/ part unslipped
Slip Not-yet-slipped
Boundary between slipped and unslipped parts
on the slip plane
Dislocation Line (One-Dimensional Defect)
Movement of an Edge Dislocation
From
W.D. Callister
Materials Science
and Engineering
Minor Answer Scripts
10-11 am and 4-5 pm from Lab
Extra ClassThursday 2nd. Nov. 10-11 am MS 702
I never did a day's work in my life. It was all fun. Thomas A. Edison
Plastic Deformation Summary Plastic deformation slip
Slip dislocations
Plastic deformation requires movement of dislocations on the slip plane
Recipe for strength?
Remove the dislocation
700
50
Stress, MPa
strainCu Whiskers tested in tension
Fig. 11.6
The critical resolved shear stress to move the dislocation depends upon
1.The width of dislocation2.The Burgers vector
WIDTH of a DISLOCATION
Narrow dislocation Wide Dislocation
Wide dislocations are easier to move than narrow dislocations
Width of a dislocation in crystal of different types of bonding:1. Covalent crystal:
Strong and directional bond: narrow dislocation
brittle2. Metallic crystal:
Weak and non directional bondswide dislocation
ductile3. Ionic Crystal
Weak and non-directional bond but large bbrittle
Eqn. 11.13
Not in course
Effect of temperature on dislocation motionHigher temperature makes the dislocation motion easier
WFe S
i
Al2O3
Ni
Cu
18-8 ssYi
e
l
d
s
t
r
e
s
s
T/Tm0 0.7
Fig. 11.8
Eqn. 11.1411.15
11.1611.17
11.18
Recipe for strength
Remove the dislocation: Possible but Impractical
Alternative:Make the dislocation motion DIFFICULT
Strengthening Mechanisms Strain hardening
Grain refinement
Solid solution hardening
Precipitation hardening
Movement of an Edge DislocationA unit slip takesplace only whenthe dislocation
comes out of thecrystal
During plastic deformation dislocation density
of a crystal should go down
Experimental Result
Dislocation Density of a crystal actually goes up
Well-annealed crystal: 1010 m-2
Lightly cold-worked: 1012 m-2
Heavily cold-worked: 1016 m-2?
Dislocation Sources
F.C. Frank and W.T. Read
Symposium on
Plastic Deformation of Crystalline SolidsPittsburgh, 1950
AB
P
Q
b
b
b
http://zig.onera.fr/~douin/index.html
b
http://zig.onera.fr/~douin/index.html
bb
Fig. 11.9
Problem 11.11
Strain Hardening or Work hardening
Strain,
y
y
During plastic deformation dislocation density increases.
Dislocations are the cause of weakness of real crystals
Thus as a result of plastic deformation the crystal should weaken.
However, plastic deformation increases the yield strength of the crystal: strain hardening or work hardening
?
Dislocation against Dislocation
A dislocation in the path of other dislocation can act as an obstacle to the motion
of the latter
Strain Hardening
]110[21
)111(
]110[21
)111(
)001(
]011[21
]110[21
]011[21
Sessile dislocation in an FCC crystal
Eqn. 11.20
222
222 aaa+Tm is the m.p. in K.
CREEP
Fig. 11.15
CreepDislocation climb
Vacancy diffusion
Cross-slip
Grain boundary sliding
Creep Mechanisms of crystalline materials
Cross-slip
In the low temperature of creep screw dislocations can cross-slip (by thermal activation) and can give rise to plastic strain [as f(t)]
Dislocation climb
Edge dislocations piled up against an obstacle can climb to another slipplane and cause plastic deformation [as f(t), in response to stress]
Rate controlling step is the diffusion of vacancies
Diffusional creep
In response to the applied stress vacancies preferentially move from surfaces/interfaces (GB) of specimen transverse to the stress axis tosurfaces/interfaces parallel to the stress axis causing elongation
This process like dislocation creep is controlled by the diffusion ofvacancies but diffusional does not require dislocations to operate
Flow of vacancies
Coble creep low T Due to GB diffusion
Nabarro-Herring creep high T lattice diffusion
Grain boundary sliding
At low temperatures the grain boundaries are stronger than the crystalinterior and impede the motion of dislocations
Being a higher energy region, the grain boundaries melt before the crystalinterior
Above the equicohesive temperature grain boundaries are weaker thangrain and slide past one another to cause plastic deformation
Creep Resistant Materials
Higher operating temperatures gives better efficiency for a heat engine
Creepresistance
Dispersion hardening ThO2 dispersed Ni (~0.9 Tm)
Solid solution strengthening
High melting point E.g. Ceramics
Single crystal / aligned (oriented) grains
Cost, fabrication ease, density etc. are other factors which determine the final choice of a material
Commonly used materials Fe, Ni, Co base alloys Precipitation hardening (instead of dispersion hardening) is not a good
method as particles coarsen (smaller particles dissolve and largerparticles grow interparticle separation )
Ni-base superalloys have Ni3(Ti,Al) precipitates which form a lowenergy interface with the matrix low driving force for coarsening
Cold work cannot be used for increasing creep resistance as recrystallization can occur which will produced strain free crystals
Fine grain size is not desirable for creep resistance grain boundary sliding can cause creep elongation / cavitation Single crystals (single crystal Ti turbine blades in gas turbine
engine have been used) Aligned / oriented polycrystals
No Dislocations
Ultra Strong Crystals
Whiskers
Composite Materials
Various Crystal Defects
Disloca-tions
Grain Boundary
G-P zone
Substitu-tionalsolute
Interstitial solute
Stacking fault
Vacancy (Diffusion)
Moral of the Story
Strength depends upon defects
Microstructure
Structural features observed under a microscope Phases and their distribution Grains and grain boundaries Twin boundaries Stacking faults Dislocations
Hierarchy of Structures
engineering structure
macrostructure
microstructure
crystal structure
atomic structure
nuclear structure
Physics and chemistry
Metallurgy and Materials Science
Engineering: Civil, Mechanical, etc. 1m
1mm
1m1nm
1A0
Real Moral of the Story
Structure Sensitive vs
Structure Insensitive Properties
For true understanding comprehension of detail is imperative. Since such
detail is well nigh infinite our knowledge is always
superficial and imperfect.
Duc Franccois de la Rochefoucald(1613-1680)