mcen2024 02-24-2011

7/17/2019 MCEN2024 02-24-2011

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7/17/2019 MCEN2024 02-24-2011


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

7/17/2019 MCEN2024 02-24-2011


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

7/17/2019 MCEN2024 02-24-2011


• 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#


7/17/2019 MCEN2024 02-24-2011


• 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!)


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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


7/17/2019 MCEN2024 02-24-2011


•

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


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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


(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

7/17/2019 MCEN2024 02-24-2011


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

7/17/2019 MCEN2024 02-24-2011


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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•

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

7/17/2019 MCEN2024 02-24-2011


• 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


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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


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•

One mechanism for dislocation lock

–

Consider two dislocations b1 and b2 in FCC


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.


b1

b2

b3

b3 is not on a slip plane

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•

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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•

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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• 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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• 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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•

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.

mcen2024 02-24-2011

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