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

Example: Slip systems in FCCSlip planes: {111} plane family in FCC possesses the highest atomic planar densitySlip directions: <110> 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).<110>: 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 <110>directions = 12 slip systems

[011]

[101]

[110]

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.

Plastic Deformation in Polycrystals

Before, undeformedequiaxial grains

The plastic deformation has produced elongated grains

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

• Concomitant applied normal stress

• Minimum stress to introduce yielding occurs when• Then

max)cos(cos λφτ

σ crssy =

crssy

o

τσφλ

245

===

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

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 psiMPaMPa

y ==σ

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

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 ?

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.

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 109

to 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 += 0

Where 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)

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

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.

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 %

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.

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.

33%CW Brass4s at 580oC 8s at 580oC

Grain Growth after 15 min; and after 10min at 700oC

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.

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.

•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

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.)

• 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.

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)

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

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