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!

!"#$%&' ) !*+,+-&.

# 

$%&'()*+(,& *,- ./01,2/31,%,2 41)3*,%&5&

 6  731 0('1 (8 -%&'()*+(,& %, 9'*&+) -18(05*+(,

 6 

./01,2/31,%,2 51)3*,%&5& %, 5*/10%*'&

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2

Plastic deformation, revisited (again)

•  When stress is

applied to induce

plasticdeformation,

atomic planes slip

•  Atomic planes slip

via the movement

of dislocationsUndeformed (no load)

Elastic deformation:

 Atomic bonds stretched

Elastic and plasticdeformation: Bonds

stretched and atomicplanes “slip”

Plastically deformed

(no load): Atomic

planes remain slipped

 A

X

Y

Z

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3

Major slip systems

BCC: 12 slip systems, 6 unique planes, 2 directions per plane

FCC: 12 slip systems, 4 unique planes, 3 directions per plane

HCP: 3 slip systems, 1 unique plane, 3 directions

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4

Comparison of FCC, BCC, and HCP

slip systems•

 

FCC and BCC both have 12 major slip

systems, while HCP only has 3

 – 

Dislocations move via slip systems anddislocation movement is the mechanism of

plastic deformation

 – 

Therefore, HCP materials are less likely to

plastically deform. i.e. HCP materials aretypically brittle

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 Atomic slipping via dislocation

motion•  Plastic deformation occurs via

the slipping of atomic planeswhich is enabled by themovement of dislocations

•  The dislocation motionmechanism of slip occurs inthe presence of shear force

•  How do dislocations moveduring a tensile test when theapplied load is axial?

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Dislocation motion in a tension testShear force IS present

in a sample in pure

tension, consider a free

body diagram of a

volume element of the

tensile sample

• “F” is the applied axial

force

• “A” is the top area of

the volume element• The stress state of this

volume element is

simply !=F/A for the

plane of “A”

F

F

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• 

We will section our volume

element to determine the

stress state of an arbitrary

plane at " from F

•  Note as we proceed that the

stress state becomes more

complex

(please review section 6.2)

Dislocation motion in a tension test

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•  From geometry, we can findthe resulting forces acting onour plane –  The normal force on the plane

is (F cos ") and the shear

force on the plane is (F sin"

) –  The area of the plane is (A/cos ")

•  Solving for the shear stress:

"  =F shear

 A plane

= F sin# 

 Acos# ( )"  =

F 

 Asin# cos# =$  sin# cos# 

Dislocation motion in a tension test

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•  From geometry, we can findthe resulting forces acting onour plane –  The normal force on the plane

is (F cos ") and the shear

force on the plane is (F sin"

) –  The area of the plane is(A/cos ")

•  Solving for the shear stress:

"  =F shear

 A plane

= F sin# 

 Acos# ( )"  =

F 

 Asin# cos# =$  sin# cos# 

Maximum value of shear stress

occurs for plane oriented at 45º(1/2!)

Dislocation motion in a tension test

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• 

Maximum shear

stress at 45º to

applied tensile force – 

For single crystals,

slip will occur on the

slip plane closest to

45º

Dislocation motion in a tension test

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• 

Maximum shear stress

at 45º to applied tensile

force

 – 

For polycrystals, slip willoccur within each grain

on the slip plane closest

to 45º

 – 

The slip path averagedacross all of the grains

will be 45º

Dislocation motion in a tension test

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• 

These rings of slip

steps on the surface

of the material are

visible in the SEMvideo of the single

crystal wire in tension

Dislocation motion in a tension test

(Show movie here.)

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 Another mechanism for plastic

deformation: twinning

• 

With some materials, theformation of twinboundaries allows plasticdeformation

 – 

NiTi shape memoryalloy is an example

• 

The amount ofdeformation by twinning issmall

• 

Twinning can reorientcrystal grains in a directionmore favorable to slip

• 

 A possible deformationpathway if slip is restricted(eg, low temps)

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Comparison: Slip versus

Twinning

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Strengthening mechanisms

• 

Since plastic deformation (yielding) occurs by

the movement of dislocations, it is possible to

strengthen a material (increase yield stress) by

preventing the motion of dislocations

•  Dislocation motion may be restricted by:

 –  Impurity atoms

 –  Increased dislocation density (strain hardening)

 –  Grain boundaries

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Strengthening by impurity atoms

•  Metals may be strengthened with adding

impurities, both substitutional and interstitial.

 –  Example shown is for Nickel in Copper. Both yield

and tensile strength are increased

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Strengthening by impurity atoms

• 

Strengthening

mechanism

 –  Impurity atoms can

reduce lattice strainaround a dislocationdue to the atomic

size difference

 –  The decreased

lattice strain leadsto the dislocation

being “pinned”

Energy lowering configurations!

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Strengthening by strain hardening

•   A ductile metal becomesstronger (increased yieldstrength) as it is plasticallydeformed: Strain hardening

 – 

This is seen in the stress-straincurve after elastic recovery

•  The increased strength is aresult of decreased  dislocation mobility caused by

increased  density ofdislocations –  How does the dislocation

density increase?

 –  Why does this restrictdislocation mobility

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Strengthening by strain hardening

• 

One mechanism for multiplication of dislocations: Frank-Read source

 –  Consider a dislocation where B and C are pinned, but B-C is mobile

 –  Dislocation loops will be generated with applied shear stress, but

not indefinitely. The dislocation loops will eventually encounter

obstacles, inducing lattice strain and preventing generation of moredislocations

http://lem.onera.fr/DisGallery/source.html

http://lem.onera.fr/DisGallery/source.html

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•  Increased dislocationdensity results indecreased dislocationmobility due to

repulsive forcebetween dislocations –  Consider the lattice

strain of an edgedislocation

 –  Two dislocations mayattract each other andcombine or they mayrepel each other

Strengthening by strain hardening

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•  Two dislocations mayattract each other andcombine or they mayrepel each other

 – 

Dislocation combinationdoes NOT always leadto annhilation

 –  Dislocations maycombine into a newdislocation that is

“locked” from motion –   A “locked” dislocation

will prevent motion ofother dislocations

Strengthening by strain hardening

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• 

One mechanism for dislocation lock

 – 

Consider two dislocations b1 and b2 in FCC

Strengthening by strain hardening

b1

b2 

b1 and b2 are moving

on {111}

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• 

 An applied shear stress would lead b1 and b2 to combineat the intersection of their slip planes, creating b3, which isin the (001) plane. This leads to a “lock” of the dislocationin a non-slip plane.

Strengthening by strain hardening

b1

b2 

b3 

b3 is not on a slip plane

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Strengthening by strain hardening

• 

Extensive strain hardening results in a

“forest of dislocations”•  http://lem.onera.fr/DisGallery/forest.html

•  Dislocation density of materials

 – 

 Annealed single crystals: 103-104

 –   Annealed polycrystalline material: 105-106

 – 

Significantly strain hardened: 109-1010

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Strengthening by strain hardening

• 

Quantifying strain hardening

 – 

Strain hardening is also referred to as cold-

working

 – 

Strain hardening is typically measured by

percent cold work

•  %CW=(A0-Ad)/A0 X 100

 –  A0 is original area

 – 

 Ad is deformed area

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• 

Materials may also

be strengthened by

the presence of grain

boundaries –  Grain boundaries

resist the intergranular

movement of

dislocations –  Example shown is a

Cu-Zn alloy (brass)

Strengthening by grain boundaries

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Strengthening by grain boundaries

•  In a single crystal grain, a dislocation moves on the slipplane with the highest shear stress

•   At a grain boundary, atomic planes between adjacentgrains are angularly misaligned

• 

When the dislocation encounters the grain boundary, themisaligned planes hinder the movement of thedislocation

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Strengthening by grain boundaries

•  Two mechanisms hinder the dislocation motion –  The slip planes may be discontinuous (offset) between the two grains

 –  The slip planes will be in different orientations, causing the dislocation tochange direction

•  For high angular misalignments, the dislocation may not traverse the

grain boundary. Instead, the dislocation density increases at the grainboundary and the resulting lattice stress creates new dislocations acrossthe boundary

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Strengthening by grain boundaries

• 

Small grain materials have more grain boundaryarea than large grain materials. Therefore, smallgrain materials resist dislocation motion more

than large grain materials and have a higheryield strength.

• 

The relationship between yield strength andgrain size may be estimated by the Hall-Petchequation:

!Y= !0 + kyd-1/2

•  where !0 and ky are material specific constants

•  this relationship is not intended for grains <1 µm or >1 mm

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 Annealing and plastic deformation

• 

Strain hardening…

 – 

creates internal lattice stress/strain from

movement and generation of dislocations

 – 

restructures crystal grains by atomic slip

•  By annealing (heat treatment), the strain

hardening effects can be reversed.