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Background Knowledge
Yield StrengthMetals “yield” when dislocations start to move
(slip). “Yield” means permanently change shape.
STRENGTHENING MECHANISM IN STRENGTHENING MECHANISM IN
METALSMETALS
Slip Systems• Slip plane: the plane on which deformation occurs,
possess the highest atomic
density.• Slip direction: the direction within the slip plane
and is always along a line of the
highest atomic density• Slip systems: a crystal deforms by
motion of a dislocation on a slip plane and in a
certain directionslip system = slip plane + slip direction
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Example: Slip systems in FCCSlip planes: {111} plane family in
FCC possesses the highest atomic planar densitySlip directions:
direction family in FCC possesses the highest atomic density
{111}: eight octahedral planes in a cube, only 4 of them need to
be considered (the other 4 are parallel planes).: total six, but,
only three lie in each of the {111} slip plane.Ex: (111) slip plane
contains the [011], [101], &[110]Therefore, 4 {111}planes x 3
directions = 12 slip systems
[011]
[101]
[110]
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Macroscopic slips in a single crystalSlips in a zinc single
crystal
Deformation of polycrystalsSlip occurs in well-defined
crystallographic planes within each grain, but more than one slip
plane is possible and likely.
In different grains, the slip planes will have different
orientations because of the random nature of the crystal
orientations.
Microscope photograph of actual shear offsets in different
grains, on surface of a copper bar.
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Plastic Deformation in Polycrystals
Before, undeformedequiaxial grains
The plastic deformation has produced elongated grains
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Slip in Single Crystals: Resolved Shear StressAngle λ: between F
& slip directionAngle φ: between F & the normal direction
of slip planeThe resolved force in slip direction Fs
Fs = F cos λThe area of the slip plane
As = A/cos φResolved shear stress
φλσφλτ coscos
cos/cos
===AF
AF
s
s
Metal single crystal – a number of potential slip planes
existsOne generally orientated most favorably – largest resolved
shear stress
( )maxmax coscos φλστ =Slip occurs when (crss = critical
resolved shear stress) CRSSττ =max
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• Concomitant applied normal stress
• Minimum stress to introduce yielding occurs when• Then
max)cos(cos λφτ
σ crssy =
crssy
o
τσφλ
245
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Example 1:Given: Single Crystal BCC iron
Tensile stress applied along [010] directionRequired:Compute the
resolved shear stress along the (110) plane and [ 11] direction
when a tensile stress of 52 MPa (7,500 psi) is applied. If slip
occurs on (110) plane and in a [ 11] direction, and resolved shear
stress is 30 MPa (4,500 psi), calculate applied tensile stress to
initiate yielding.
1
1
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Solution:
φ angle between (110) plane normal and the [010] direction is
450
From triangle ABC λ= tan-1 = 54.70
τR = σ cosλ cosφ = (52MPa)(cos 45)(cos 54.7) = 21.3 MPa
(3,060psi)
)/2( aa
)600,10(4.73)7.54(cos)45(cos
3000 psiMPa
MPay ==σ
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Example 2Problem: A FCC crystal yields under a normal stress of
2MPa applied in the [123] direction. The slip plane is (111) &
slip direction is [101]. Determine critical resolved shear
stress.Solution:
MPayc 933.0756.0617.02coscos
756.0214
311)1(32)1(
]101[]321[cos
617.0314
32111132)1(
)111(]321[cos
22222
222222
=××==
=+
=+−++−
−•−=
=++−
=++++−
•−=
φλστ
λ
φ
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Mechanisms of Strengthening• The ability of a metal to
plastically deform depends on the ability of
dislocations to move• Reducing or inhibiting mobility of
dislocations enhances mechanical
strength
These can be used for increasing the material strength, but
ductility may be lost.
Dislocation motion may be inhibited by:
Point defect (solution hardening)
Other dislocations (entangling)Grain boundaries Dislocation
forest
Other phases - Precipitates
When Slip is Inhibited ?
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Strengthening of MetalsThere are 4 major ways to strengthen
metals, and all work because they make dislocation motion more
difficult. They also reduce the ductility:1)Cold work (Strain
Hardening)2)Reduce grain size (Strengthening by Grain Size
Reduction)3)Add other elements in solid solution (Solid Solution
Strengthening)4)Add second phase particles (Precipitation or Age
Hardening)
• These mechanisms may be combined. • For example, the world’s
strongest structural material (with some ductility) is steel piano
wire. It combines all four strengthening mechanisms, and can have a
yield strength of 500,000 psi. One wire, 0.1” in diameter, can hold
up a 4,000 lb Ford Explorer.
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STRAIN HARDENING• Ductile material becomes harder and stronger
as it is plastically deformed• The dislocation density – expressed
as total number dislocation length per unit
volume – mm/mm3 increases from 105 to 106 mm-2 for a heat
treated metal to 109to 1010 mm-2 for a heavily deformed metal.–
Dislocation strain field interactions– Dislocation density
increases with deformation or cold working– Dislocations are
positioned closer together– On average, dislocation-dislocation
strain fields are repulsive
dislflow k ρττ += 0Where ρdisl: dislocation density
nTT K εσ =
n = strain hardening exponent – measures the ability of a metal
to harden
n ~ 0.5 (FCC) n ~ 0.2 (BCC)n ~ 0.05 (HCP)
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τcrss versus densityCold Working
Cold working: plastic deformation of a metal or alloy at a
temperature where dislocations are created faster than they are
annihilated
100%0
0 ×⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
AAACW d
Where, %CW: percent of cold workA0: original cross-sectional
areaAd: area after deformation
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Influence of Cold Working on Mechanical Properties
As the yield strength and the tensile strength increases with
increasing amount of cold working, the ductility of the metal
decreases.
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Example:Given : Copper rod is cold worked such that its diameter
is reduced from 15.2 mm to 12.2 mm.
Determine its tensile strength and ductility
%6.35100
22.15
22.12
22.15
% 2
22
=
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛
= xCWπ
ππ
From Figure TS vs CW TS = 340 MPa From Figure ductility vs CW %
EL = 7 %
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Effects of plastically deformed polycrystalline metal at
temperature less than Tm:•Change in grain shape•Strain
hardening•Increased dislocation density•Stored energyWhen metals
are plastically deformed about 5% of deformation energy is retained
internally associated with dislocations. The properties of the cold
worked metal (partially or totally) can be restored by:Recovery
and/or Recrystallization and Grain Growth
RecoverySome of the stored internal strain energy is relieved by
virtue of dislocation motion as a result of enhanced atomic
diffusion at elevated temperature.Effects of recovery in cold
worked metals:•Ductility increases•Yield and tensile strength
decreases slightly•Hardness decreases slightly.•Metal toughness
increases.•Electrical and thermal conductivity of the metal is
recovered to their precold-worked states.•There is no apparent
change in the microstructure of the deformed material.
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Recrystallization
Cold worked material• high dislocation density• lot of stored
energy• very strong• not very ductile
Recrystallized material• low dislocation density• no stored
energy• weak• ductile
process
•After recovery – grains remain at relatively high energy
states•Recrystallization – formation of a new set of strain-free
and equiaxed grains, low dislocation densities•Driving force –
difference in internal energy between strained and unstrained
material•New grains form as small nuclei – grow and replace parent
material – short –range diffusion• The process is a heat treating
process called annealing. Annealing requires high temperature.
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33%CW Brass4s at 580oC 8s at 580oC
Grain Growth after 15 min; and after 10min at 700oC
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Recrystallization Temperature•Temperature at which
recrystallization just reaches completion in one hour.•450oC for
the above example.• Typically between 0.5 to 0.33 the melting point
of the metal.•Depends on the amount of cold work and of the
impurity level of the alloy.•There is a critical degree of cold
work below which recrystallization can not be made to occur.
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Notes on Recrystallization:• The amount of cold work controls
the initial recrystallized grain size. More
cold work ® more stored energy → easier nucleation → more
nucleation sites → smaller grain size.
• The temperature and time of annealing controls the final grain
size, if there is substantial growth after recrystallization. Grain
growth requires diffusion, and diffusion is faster at higher
temperatures. The time at temperature controls the total amount of
diffusion.
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•A fine grain size has many benefits beyond strength. •In
general, finer grain sizes are more resistant to fatigue and
fracture failures, and have more reproducible and homogeneous
mechanical properties. •Finally, in general, metals with fine grain
size are also more easily formed in metalworking operations than
metals with coarse grain sizes.
Grain Growth• Grains continue to grow following
recrystallization at elevated temperatures• Energy is reduced as
grains grow in size• As large grains grow – small grains shrink•
Boundary motion – short-range diffusion of atoms from one side of
the boundary to the other.
• At a constant temperature
d0 – initial grain diameter at time (t) = 0K and n are time
independent constantsn is generally ~2
Ktdd nn =− 0
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STRENGTHENING BY GRAIN SIZE REDUCTION
• Dislocations cannot penetrate grain boundaries, because the
crystal planes are discontinuous at the grain boundaries.
• Therefore, making a smaller grain sizeincreases strength (more
obstacles and shorter mean slip distance.)
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• This can be quantified by the so-called “Hall-Petch
Equation:
• where σy is the yield strength, d is the grain size, and σo
and ky are material constants.
• The increases in strength at very small grain sizes can be
enormous. One are of current research is on so-called
“nanostructured metals”, which have grain sizes from 20 to 200 nm.
They can have very high strength.
Influence of grain size on yield strength (brass)
As d⇓, σys⇑ and the ductility or ⇑ or it is constant
5 μm100 μm
Strength triples as grain size goes from 100 μm to 5 μm.
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SOLID SOLUTION STRENGTHENING• Impurity atoms that go into solid
solution impose lattice strains on surrounding host atoms
• Lattice strain field interactions between dislocations and
impurity atoms result in restriction of dislocation movement
• This is one of the most powerful reasons to make alloys, which
have higher strength than pure metals.
• Example: 24k gold is too soft. If we put in 16% silver and 9%
copper, we get an alloy that looks just like pure gold, but is much
more strong and durable. We call this 18k gold.
(18/24 = 75% gold)
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Why it works
Big atoms (or interstitials) like to live here (there is more
space.)
Atoms of either type diffuse to dislocations during high
temperature processing, then exert forces on the dislocation later
to keep them stuck.
Small atoms like to live here. (they reduce lattice strain
caused by the dislocation).
• Small impurity atoms exert tensile strains (see figure
below)
• Large impurity atoms exert compressive strains
• Solute atoms tend to diffuse and segregate around dislocations
to reduce overall strain energy – cancel some of the strain in the
lattice due to the dislocations
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Compressive Strains Imposed by Larger SubstitutionalAtoms
Add nickel to copper, strength goes up, ductility goes down -
for the same reason: dislocation mobility is decreased. Note:
trade-off in properties