Plasticity - 2013

7/28/2019 Plasticity - 2013

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PLASTICITY

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Dislocations and Materials Classes

• Covalent Ceramics (Si, diamond): Motion hard.-directional (angular) bonding

• Ionic Ceramics (NaCl): Motion hard.

-need to avoid ++ and - -neighbors.

+ + + + + + +

+ + + +

- - - - - - -

- - -

• Metals: Disl. motion easier. -non-directional bonding-close-packed directions

for slip. electron cloud ion cores

+ +

+ +

+ + + + + + + + + + + + +

+ + + + + + +

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Dislocation MotionDislocations and plastic deformation• Cubic & hexagonal metals - plastic

deformation by plastic shear or slip whereone plane of atoms slides over adjacentplane by defect motion (dislocations).

• If dislocations don't move,deformation doesn't occur!

Adapted from Fig. 7.1,

Callister 7e.

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Dislocation Motion• Dislocation moves along slip plane in slip

direction perpendicular to dislocation line• Slip direction same direction as Burgers

vector

Edge dislocation

Screw dislocation

Adapted from Fig. 7.2,Callister 7e.

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If a material is subjected to a load of sufficient magnitude it shows permanent(irrecoverable) deformation.It is result of the permanent displacement of atoms and molecules from their originalposition.

If the deformation is continuously increasingthen the phenemenon is called FLOW.

Slip plane

P

x’

x

P P n

P s

Slipdirection

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Slip System: Slip plane anddirection• Slip plane - plane allowing easiest

slippage- Highest planar densities

• Slip direction - direction of movement - Highest linear densities

Deformation Mechanisms

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FCC

FCC Slip occurs on {111} planes (close-packed) in<110> directions (close-packed)

=> total of 12 slip systems in FCC

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(111) (111) (111) (111)

(111)

Parallel

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(111), [101] or [101]--- 1

(111), [110] or [110]--- 2

(111), [011] or [011]--- 3

1

2

3

So, for (111) plane:

Therefore, for an FCC structure:{111} - <110> , there are 12 slip systems.

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To sum up:

Slip phenemenon is used to explain the plasticbehaviour of materials.Slip occurs along certain crystal planes anddirections.Slip planes & slip directions make slip systems.For an FCC structure {111} - <110> , there are 12slip systems.

For an BCC structure {110} - <111> → 12 slipsystems.For an HCP structure → 3 slip systems.

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The stress that initiates slip is known as thecritical resolved shear stress.

Ø: Angle between thenormal to the slip planeand the applied stressdirections.

λ: Angle between the slipand stress directions.

A s

A

F

τ R = σ . cos λ . cos Ø

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Stress and Dislocation Motion• Crystals slip due to a resolved shear stress, t R .

• Applied tension can produce such a stress.

slip plane

normal, ns

Resolved shear stress: t R

= F s / A s

AS

tR

t R

F S

Relation betweens and t R

t R = F S / AS

F cos l A /cos f

l F

F S

f nS

AS

A

Applied tensilestress: = F / A s

F

A

F

flst coscosR

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Fs= F cos λ (shear force along the slip direction)

τR =

σcos

λcos Ø

A s = A / cos Ø (shearing area)

cosλ

. cos ØF

A A / cos ØF cos λ Fs

A s τ

R = = =

σ

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Single Crystal Slip

Adapted from Fig. 7.8, Callister 7e.

Adapted from Fig.7.9, Callister 7e.

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Ex: Deformation of single crystal

So the applied stress of 65 MPa will not

cause the crystal to yield.

MPa65

coscos

s

f l s t

l =35 ° f =60 °

MPa30 MPa62.26

)41.0(MPa)65(

)60)(cos35cos(MPa)65(

crsst t

t

t crss = 30 MPa

a) Will the single crystal yield?

b) If not, what stress is needed?

s = 65 MPa

Adapted fromFig. 7.7,Callister 7e.

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Ex: Deformation of single crystal

MPa25.7341.0

MPa30coscos

crss

f l t

s y

What stress is necessary (i.e., what is the yieldstress, s y )? (!We will learn about yield stress later on!)

)41.0(coscosMPa30crss y y s f l s t

MPa25.73 ys s

So for deformation to occur the applied stress mustbe greater than or equal to the yield stress

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Ex: Consider a single crystal of BCC ironoriented such that a tensile stress is appliedalong [010] direction.

a) Compute the resolved shear stress along a(110) plane and in a [111] direction when atensile stress of 52 MPa is applied.

b) If slip occurs for the above plane and

direction, and the critical resolved shearstress is 30 MPa, calculate the magnitude of the applied tensile stress necessary toinitiate yielding.

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a) Ø=45 ° angle b/w normal[110] & stress [010]

σ= 52MPa(110)

[111]z

y x

σ= 52

cos Ø =1*0 + 1*1 + 0*0

(1 2+1 2+0 2) (0 2+1 2+0 2)= 1

√2 Ø = 45 °

λ: angle b/w slip direction [111] & stress direction [010]

cos λ = -1*0 + 1*1 + 1*0(1 2+1 2+1 2) (1)

= 1√3 λ = 54.7 °

τ

R = 52 * cos 45 * cos 54.7 = 21.3 MPa

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b) σy = τ CR

cos Ø cos λ = 30

cos 45 cos 54.7

σy = 73.4 MPa

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• Stronger - grain boundariespin deformations

• Slip planes & directions (l , f ) change from onecrystal to another.

• t R will vary from onecrystal to another.

• The crystal with the largest t R yields first.

• Other (less favorably oriented) crystalsyield later.

Adapted from Fig.7.10, Callister 7e. (Fig. 7.10 iscourtesy of C.

Brady, NationalBureau of Standards [now theNational Institute of Standards andTechnology,Gaithersburg, MD].)

Slip Motion in Polycrystalss

300 mm

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• Condition for dislocation motion: CRSStt R

• Crystal orientation can make it easy or hard to move dislocation

10 -4 GPa to 10 -2 GPa

typically

flst coscosR

Critical Resolved Shear Stress

t maximum at l = f = 45º

t R = 0

l =90°

s

t R = s /2

l =45° f =45°

s

t R = 0

f =90°

s

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DISLOCATIONS&

σ - ε CURVES

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I IIIII

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Elastic Region: σy is the stress required to startplastic deformation. It is called the yield stress.From σy & on → plastic deformation starts.

Stage I: dσ /d ε ≈ 0 (work hardening rate) is

low because only the primary slip systems areactive. The slip planes are parallel to eachother and only these parallel planes will slipand they do not intersect themselves, i.e.dislocations are moving along parallel planes.

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Stage II: Other dislocations will start to movealong intersecting planes (more than one slip

system becomes active). Therefore they formbarriers to one another’s motion. It becomesharder to further deform the material. This

stage is known as Work Hardening Stage .

Stage III: The geometry of the planes have

so changed that the planes will acceleratelyslip and failure will occur.

Plasticity - 2013

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