1
Engineering materials
Lecture 2
Imperfections and defects
Response of materials to stress
2
Crystalline Imperfections (4.4)
No crystal is perfect.
Imperfections affect mechanical
properties, chemical properties and
electrical properties.
• Imperfections can be classified as
•
•
¾ Zero dimension point deffects.
¾ One dimension / line deffects
(dislocations).
¾ Two dimension deffects (grain boundaries).
¾ Three dimension deffects (voids).
3
Point Defects – Vacancy
Vacancy is formed due to a missing atom.
Vacancy is formed (one in 10000 atoms)
during crystallization or mobility of atoms.
• Energy of formation is 1 ev.
• Mobility of vacancy results in cluster of
vacancies.
• Also caused due
to plastic defor-mation, rapid
cooling or particle
bombardment.
•
•
4
Point Defects - Interstitial
• Atom in a crystal, sometimes, occupies
interstitial site.
• This does not occur naturally and can be
induced by irradiation.
• This defects caused structural distortion.
5
Interstitial Solid Solution
• Solute atoms fit in between the voids (interstices) of
solvent atoms.
• Solvent atoms in this case should be much larger than
solute atoms.
• Example:- between 912 and 13940C, interstitial solid
solution of carbon in γ iron
(FCC) is formed.
• A maximum of 2.8%
of carbon can dissolve
interstitially in iron.
Iron atoms r00.129nm
Carbon atoms r=0.075nm
6
Substitutional Solid Solution
Solute atoms substitute for parent solvent atom in a
crystal lattice.
• The structure remains unchanged.
• Lattice might get slightly distorted due to change in
diameter of the atoms.
• Solute percentage in solvent
can vary from fraction of a
percentage to 100%
•
Solvent atoms
4-15
Solute atoms
7
Line Defects – (Dislocations)
Lattice distortions are centered around a
line.
• Formed during
•
¾
Solidification
¾ Permanent Deformation
¾ Vacancy condensation
•
Different types of line defects are
¾
Edge dislocation
¾ Screw dislocation
¾ Mixed dislocation
8
Edge Dislocation
•
Created by insertion of extra half planes of atoms.
•
Positive edge dislocation
•
Negative edge dislocation
•
Burgers vector b
Shows displacement of
atoms (slip).
Figure 4.18
Burgers vector
9
Screw Dislocation
Created due to shear stresses applied to regions
of a perfect crystal separated by cutting plane.
• Distortion of lattice in form of a spiral ramp.
• Burgers vector is parallel to dislocation line.
•
10
Mixed Dislocation
•
Most crystal have components
of both edge and screw
dislocation.
•
Dislocation, since have
irregular atomic arrangement
will appear as dark lines
when observed in electron
microscope.
Dislocation structure of iron deformed
14% at –1950C
11
Grain Boundaries
Grain boundaries separate grains.
Formed due to simultaneously growing crystals
meeting each other.
• Width = 2-5 atomic diameters.
• Some atoms in grain boundaries have higher energy.
• Restrict plastic flow and prevent dislocation
movement.
•
•
3D view of
grains
Grain Boundaries
In 1018 steel
12
Planar Defects
• Grain boundaries, twins, low/high angle
boundaries, twists and stacking faults
• Free surface is also a defect : Bonded
to atoms on only one side and hence
has higher state of energy
Highly
reactive
• Nanomaterials have small clusters of
atoms and hence are highly reactive.
13
Twin Boundaries
• Twin: A region in which mirror image pf
structure exists across a boundary.
• Formed during plastic deformation and
recrystallization.
• Strengthens the metal.
Twin
Plane
Twin
14
Other Planar Defects
• Small angle tilt boundary: Array of edge
dislocations tilts two regions of a crystal by
< 100
• Stacking faults: Piling up faults during
recrystallization due to collapsing.
¾ Example: ABCABAACBABC
FCC fault
• Volume defects: Cluster of point defects
join to form 3-D void.
15
Material Properties
• Anisotropy (非均向性):
The measured material properties are dependent of
the direction and are functions of the symmetry of crystal
structure.
• Isotropy (均向性):
The measured material properties are independent
of the direction.
• Homogeneity (均質性):
The measured material properties are independent
of the position.
• In-homogeneity (非均質性):
The measured material properties are dependent of
the position.
16
Stress
Stress
σij
ΔF
= ΔA→0 ΔA
lim
– Normal stress
σ11, σ22, σ33
– Shear stress
σ12, σ13, σ23
⎡σ 11 σ 12 σ 13 ⎤
⎡⎣σ ij ⎤⎦ = ⎢σ 12 σ 22 σ 23 ⎥
⎢
⎥
⎢⎣σ 13 σ 23 σ 33 ⎥⎦
Symmetric
17
Strain
Strain:
– Normal strain
ε11, ε22, ε33
– Shear strain
ε12, ε13, ε23
⎡ε11 ε12 ε13 ⎤
⎡⎣ε ij ⎤⎦ = ⎢ε12 ε 22 ε 23 ⎥
⎢
⎥
⎢⎣ε13 ε 23 ε 33 ⎥⎦
Symmetric
1 ⎛ ∂ui ∂u j ⎞
+
ε ij = ⎜
⎟⎟
⎜
2 ⎝ ∂x j ∂xi ⎠
u1, u2, u3 displacement
18
Stress and Strain
•
Metals (materials) undergo deformation under
uniaxial tensile force.
•
Elastic deformation: Metal (material) returns to its
original dimension after tensile force is removed.
•
Plastic deformation: The metal (material) is deformed
to such an extent such that it cannot return
to its original dimension
6-10
19
Linear elasticity
• Linearity: linear behavior between input
(force) and output (displacement).
• Elasticity: elastic behavior
– Consider a body that is loaded by changes in
applied forces, displacements or temperatures.
If removal of the loading returns the body
immediately to its initial state, the material
response is called elastic.
20
Typical stress-strain curves
Ref: Landis, 2001.
21
Uniform tension or compression
For simple tension, the tensile strain is
proportional to the tensile stress
σ11 = σ,
σ22 = σ33 = σ12 = σ13 = σ23 = 0
σ= E ε
E: Young’s modulus (elastic modulus)
22
Mechanical Properties
•
Modulus of elasticity (E) : Stress and strain are
linearly related in elastic region. (Hooks law)
σ (Stress)
E=
ε (Strain)
Δσ
E=
Δε
Δσ
Strain
Δε
•
Higher the bonding strength,
higher is the modulus of elasticity.
Stress
Linear portion of the
stress strain curve
• Examples: Modulus of Elasticity of steel is 207 Gpa.
Modulus of elasticity of Aluminum is 76Gpa
23
Engineering Stress and Strain
Engineering stress σ =
(Nominal stress)
Δl
A0
F (uniaxial tensile force)
A0 (Original cross-sectional area)
Units of Stress are PSI or N/M2 (Pascals)
1 PSI = 6.89 x 103 Pa
0
0
A
Change in length
Engineering strain = ε =
Original length
(Nominal strain)
=
−
0
=
0
Units of strain are in/in or m/m.
Δ
0
24
Pure shear
σ12 = σ21 = τ,
σ11 = σ22 = σ33 = σ13 = σ23 = 0
τ=Gγ
G: shear modulus
∂u x ∂u y
Shear angle γ =
+
∂y
∂x
Ref: Lai, 1993.
25
Shear Stress and Shear Strain
Shear stress
S (Shear force)
τ=
A (Area of shear force application)
Shear strain
Amount of shear displacement
γ =
Distance ‘h’ over which shear acts.
Elastic Modulus G =
τ/γ
26
Poisson’s Ratio
Poisons ratio =
ε (lateral )
εy
ν =−
=−
ε (longitudinal )
εz
Theoretical value
−1 ≤ ν ≤ 0.5
w0
0
w
Usually poisons ratio ranges from
0 to 0.5
Example: Stainless steel
Copper
0.28
0.33
27
Hydrostatic pressure
p
σ11 = σ22 = σ33 = p,
σ12 = σ13 = σ23 = 0
Dilatation:
V − V0
Δ=
V0
p
p
Δ = (1 + ε11 )(1 + ε 22 )(1 + ε 33 ) − 1
Δ = ε11 + ε 22 + ε 33
Pressure-Dilatation relationship:
p = − KD
Κ: bulk modulus
p
Hydrostatic
pressure
28
Relations among Material’s Properties
Among the material parameters: E, G, κ and ν,
only two of them are independent.
9 KG
E=
3K + G
K=
E
3 (1 − 2ν )
-1 < ν ≤ 0.5
3K − 2G
ν=
2 ( 3K + G )
E
G=
2 (1 +ν )
29
Response of Material to Stress
For isotropic and linear
elastic materials,
1
[σ 11 −ν (σ 22 + σ 33 )]
E
1
ε 22 = [σ 22 −ν (σ 11 + σ 33 )]
E
1
ε 33 = [σ 33 −ν (σ 11 + σ 22 ) ]
E
1
ε12 =
σ 12
2G
1
ε13 =
σ 13
2G
1
ε 23 =
σ 23
2G
ε11 =
For example, hydrostatic pressure
σ11 = σ22 = σ33 = -p,
σ12 = σ13 = σ23 = 0
ε11 = ε22 = ε33 =
-(1-2ν)p
E
ε12 = ε13 = ε23 = 0
Dilatation: Δ = ε11 + ε 22 + ε 33
=−
K =−
3(1 − 2ν ) p
E
p
E
=
Δ 3 (1 − 2ν )
30
Shear distortion
•
Shear deformation only
changes shape when
deformation is small
•
Infinitesimal shear does
not change volume
31
Tensile test
Strength of materials can be tested by pulling the metal to failure.
Load Cell
Specimen
Extensometer
Force data is obtained from Load cell
Strain data is obtained from Extensometer.
6-14
32
Engineering Stress and Strain
Engineering stress σ =
(Nominal stress)
Δl
A0
F (uniaxial tensile force)
A0 (Original cross-sectional area)
Units of Stress are PSI or N/M2 (Pascals)
1 PSI = 6.89 x 103 Pa
0
0
A
Change in length
Engineering strain = ε =
Original length
(Nominal strain)
=
−
0
=
0
Units of strain are in/in or m/m.
Δ
0
33
Tension tests
Load-displacement
Stress-strain
34
Tensile Test (Cont)
σUTS
Figure 5.21
Commonly used
Test specimen
Typical Stress-strain
curve
6-15
35
Yield Strength
•
•
•
Yield strength is strength at which
metal or alloy show significant
amount of plastic deformation. σ0.2%
(or 0.2% proof stress)
0.2% offset yield strength is that
strength at which 0.2% plastic
deformation takes place.
Construction line, starting at 0.2%
strain and parallel to elastic region
is drawn to fiend 0.2% offset yield
strength.
36
Ultimate tensile strength
Ultimate tensile strength (UTS) is the maximum
strength reached by the engineering stress strain
curve.
Al 2024-Tempered
• Necking starts after UTS is
S
reached.
T
•
R
E
S
S
M
p
a
• More ductile the metal is, more
is the necking before failure.
• Stress increases till failure. Drop
in stress strain curve is due to stress
calculation based on original area.
6-19
Necking Point
Al 2024-Annealed
Strain
37
Percent Elongation
Percent elongation is a measure of ductility of a
material.
• It is the elongation of the metal before fracture
expressed as percentage of original length.
•
Final length – initial Length
% Elongation =
Initial Length
Measured using a caliper fitting the fractured
metal together.
• Example:- Percent elongation of pure aluminum is
35%
For 7076-T6 aluminum alloy it is 11%
•
6-20
38
Percent Reduction in Area
•
•
Percent reduction area is also a measure of ductility.
The diameter of fractured
end of specimen is measured using caliper.
% Reduction =
Area
Initial area – Final area
initial area
• Percent reduction in area
in metals decreases in case
of presence of porosity.
6-21
Stress-strain curves of different metals
39
Stress-strain curve
•
During plastic flow, the total volume of the specimen is conserved: matter
is just flowing from place to place.
σUTS
E
εf
40
Strain Energy
The strain energy of a deformed material per unit volume is given by the total
area under the stress-strain curve
ε*
U = ∫ σ dε
0
Modulus of Resilience
σy 2
Ur=
2E
=elastic strain energy + plastic strain energy
41
True Stress – True Strain
•
True stress and true strain are based upon instantaneous
cross-sectional area and length.
• True Stress = σt =
F
Ai (instantaneous area)
i
d
li
A0
= ln = ln
l0
Ai
∫
• True Strain = εt =
• True stress is always greater than engineering stress.
0
42
True Stress-True Strain for Plastic Flow
During plastic flow, the total volume of the specimen is conserved:
matter is just flowing from place to place.
Assume volume is constant
AoLo=AL
Ao
L
=
= 1+ ε
A Lo
σt =
L
A
F F Ao
=
= σ o = σ (1 + ε )
A Ao A
A
ε t = ∫ dε t =
L0
L
∫
L0
d
= ln
A
L
= ln 0 = ln(1 + ε )
L0
A
43
Necking in True Stress-True Strain Curve
dσ
=0
dε
Necking starts at
Where the ultimate strength occurs in the engineering stress-strain curve
That is at
F=σt A
F = Fmax
dF =0
dF = σ t dA + Adσ t = 0
dσ t
σt
=−
dA dL
=
= dε t
A
L
dσ t
= σt
dε t
is the condition for necking in true
stress-stain curve
44
Hardness and Hardness Testing
Hardness is a measure of the resistance of a metal to
permanent (plastic) deformation. Rockwell hardness
tester
• General procedure:
•
Press the indenter that
is harder than the metal
Into metal surface.
Withdraw the indenter
Measure hardness by
measuring depth or
width of indentation.
Figure 6.27
45
Hardness Tests
Hardness=
Force
Total surface
area of indent
46
Hardness
• True Hardness H = F/Aprojected = 3 σy
• Vicker’s Hardness : Hr = F/Ainside surface area of
the indent
47
Plastic Deformation in Single Crystals
Plastic deformation of single crystal results in step
markings on surface
slip bands.
• Atoms on specific crystallographic planes (slip
planes) slip to cause
slip bands.
•
Slip bands of zinc
Figure 6.28
48
Slip Bands and Slip Planes
Slip bands in ductile metals are uniform (occurs in
many slip planes).
• Slip occurs in many
Slip lines
slip planes within
slip bands.
• Slip planes are
about 200A thick
and are offset by
about 2000A
•
Figure 6.30
Slip bands
49
Plastic Deformation by the Slip Mechanism
50
Slip Mechanism
•
•
During shear, atoms do not slide over each other simultaneously.
The slip occurs due to movement of dislocations.
Wall of high dislocation density
Dislocation density :
Sum of length of dislocation line per
unit volume
1
ρ = ∫ ds
V
ρ=1010~1012 cm/cm3
for highly deformed crystals
Dislocation cell structure in lightly
deformed Aluminum
51
Slip in Crystals
Slip occurs in densely or close packed planes.
Lower shear stress is required for slip to occur in
densely packed planes.
• If slip on close packed
Close packed
planes is restricted ,
plane
then less dense planes
become operative.
• Less energy is required
to move atoms along
denser planes.
•
•
6-32
Non-close-packed
plane
52
Slip Systems
Slip systems are combination of slip planes and slip
direction.
• Each crystal has a number of characteristic slip
systems.
• In FCC crystal, slip takes place in {111} octahedral
planes and <110> directions.
•
4 (111) type planes and 3 [110] type directions.
4 x 3 = 12 slip systems.
Table 5.3
6-33
53
Slip Planes and Directions for FCC
Crystals
54
Slip Systems in BCC Crystal
•
BCC crystals are not close packed. The slip
predominantly occurs in {110} planes that has
highest atomic density.
55
Glide force f
Glide force f is the force
acting on a dislocation
External work = (τ L1 L2 )b
Internal work = (f L1) L2
Equivalence between external
and internal work: f = τ b
Glide force f is defined per unit
length of dislocation line
56
Critical Resolved Shear Stress
Critical resolved shear stress is the stress required to
cause slip in pure metal single crystal.
• Depends upon
•
¾
¾
¾
¾
Crystal Structure
Atomic bonding characteristics
Temperature
Orientation of slip planes relative to shear stress
Slip begins when shear stress in slip plane in slip
direction reaches critical resolved shear stress.
• This is equivalent to yield stress.
• Example :- Zn
HCP 99.999% pure
0.18MPa
Ti
HCP 99.99% pure
13.7 MPa
Ti
HCP
99.9% pure
90.1 MPa
•
57
Schmid’s Law
•
The relationship between uniaxial stress action on a
single cylinder of pure metal single crystal and
resulting resolved shear stress produced on a slip
system is give by
Normal to
τr=
Slip plane
Shear Force
Shear Area
Fr
F cos Φ
F
=
=
=
cos λ cos Φ
A1 A0 / cos λ A0
= σ cos λ cos Φ
λ
Slip
directi
A1=Area of
Slip plane
σ=
6-37
Φ
F
A0
τr =
Fr
A1
58
Example
Calculate the resolved shear stress on the (111)[011] slip
system of an unit cell in a FCC nickel single crystal if a stress
of 13.7 MPa is applied in the [001]direction
(0, −1,1) • (0, 0,1)
1
2
02 + 12 + 12 02 + 02 + 12
(1,1,1) • (0, 0,1)
1
cos φ =
=
3
12 + 12 + 12 02 + 02 + 12
1
1
×
= 5.6 MPa
τ r = 13.7 ×
2
3
cos λ =
=
59
Theoretical Shear Strength (Prediction)
dhkl
2π x
τ = k sin
b
for small deformation
2π x and τ = Gγ = G x
τ ≈k
d hkl
b
assume
2π
G
k=
b
d hkl
τ max
k=
Gb
2π d hkl
Gb τ max 1
1
≈ ~
=
10 20
2π d hkl G
60
Theoretical Tensile Strength (Prediction)
r-ro
ε= r
o
σ=NF=F/ro2
Interatomic bonding force
σ
2πε ⎞
⎟
⎝ λ ⎠
2πε ⎞
⎟
⎝ λ ⎠
σ = k sin ⎛⎜
σ = k sin ⎛⎜
ε
ε=0
assume σ
dσ
dε
=
ε =0
2π k
λ
= Eo
k=
Eo
for λ=2
π
σ max ≈
Eo
π
σ max
⎛ 2πε ⎞
= k sin ⎜
⎟
⎝ λ ⎠
Interatomic force typically is
negligible at r=2r0
1
1
≈
∼
~
E
10 20
61
Uniaxial Tension Test of Polycrystals
F
Shear stress τ
θ
τ=
F sin θ
F
= sin θ cos θ = σ sin θ cos θ
A / cos θ A
Maximum shear stress occurs at θ = 45o
τ max =
F
Crosssection A
σ
2
So that yielding occurs at σ y = 2τ y