MSE 2034 Dislocations and Strengthening Mechanisms

7/29/2019 MSE 2034 Dislocations and Strengthening Mechanisms

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Theoretical strength of

crystalsis much greater

then their experimentalstrength

Dislocations and Strengthening

Mechanisms


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g

Plastic deformationcorresponds to the

motion of large

numbers ofdislocations


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dislocation density:

total dislocation length/unitvolume

number of dislocations/unit area

Carefully prepared crystal: 103/mm2

Heavily deformed crystal: 109/mm2

After heat treatment: 106/mm2


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Metals under plastic deformation emit

energy

95% heat

5% internal energy stored as strain


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Shear Stress


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gamma-irradiated LiF single crystal.
http://upload.wikimedia.org/wikipedia/commons/0/05/Dislocation_pile-up.jpg


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Edge dislocation

Screw dislocation


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Edge dislocation


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Slip: process by which plastic deformation isproduced by a dislocation motion.

Slip plane: crystal plane along which thedislocation travels

Slip direction: direction of motion

Slip System = slip plane + slip

direction


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planeslipofarea

directionslipinforcetorque

F l

F cos l

planeslipofarea

F l

cos


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A2

planeslipofarea

F l

cos

A1

fphi

1

2cosA

Af

A2

A1 fcos2

1

AA

f

l

cos

cos2A

F22

coscos

cos

cosA

FA

F fl

f

l lfl

f

l coscoscoscos

cos

cos

22

A

FA

F


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fl coscos

(111)

FCC

xz

y]110[

]011[

]110[


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Slip Systems

Cu ductility of 45% elongation

Ti ductility of 25% elongation


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Zn


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Resolved shear stress

The most favorable slip system

fl coscosR

max)cos(cos(max) fl

R


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max)cos(cos(max) fl R

Critically resolved shearstress: Minimum shear stress needed to start slip

This occurs at yield

max)cos(cos lf crssy

max)cos(cos flYcrss


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max)cos(cos(max) fl R

Critically resolved shearstress: Minimum shear stress needed to start slip

This occurs at yield

max)cos(cos flYcrss

(max)Rcrss


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Minimum stress appears

atf=l=45

cos (45)= 1/2y2crss

fl coscosYcrss


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A single crystal of cadmium is oriented for a tensile test so its slip

plane makes an angle of 65 with the tensile axis. Possible slip

directions are 30, 48, 78.

a) Which of the slip directions is most favored?b) If plastic deformation begins at a tensile stress of 1.55 MPa, find

crss

a) cos 30= 0.87, cos 48= 0.67, cos 78=0.21

l= 30 most favored

b)

fl coscosYcrss

65

MPaMPacrss 57.065cos30cos55.1

At =1.0 MPa, will the system slip?

No


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Slip lines on copper


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Polycrystalline material

Before deformation afterdeformation

Equiaxial grains elongated grains


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Deformation by twinning

(mechanical)

Slip

twinning


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Strengthening in metals

Grain size reductionSolid-solution

Strain hardening


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Grain size reduction

Grain Size is regulated by

1. Rate of solidification2. Plastic deformation

3. Heat treatments


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Grain size reduction

Grain Boundary

barriers

Grain orientation

differentAtomic disorder at

boundaries


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Hall-Petch

Equation

y = 0 + kyd-.5

Grain sizereduction


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Strain hardening


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Solid-solution- Ni and Cu


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Solid-solution Mechanism

Foreign atoms host atoms

compressive

lattice strain


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Strain hardening


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Strain hardening

Work hardening Cold working

Process in which a ductile metal becomes harder

and stronger as it is plastically deformed.

Ad area after deformation

100%0

0

A

AACW d


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Strain hardening


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RecoveryRecrystallization

Grain Growth

Heat

Treatment

(annealing)


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Recovery Recrystallization Grain Growth

Recovery

Stored internal strain

energy relieved by

dislocation motion


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Recrystallization

Formation of a new set

of strain-free crystals

within a previously cold

worked material.


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Grain Growth

Increase in average

grain size of a

polycrystalline material.


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33% cold worked brass


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Recrystallization 3s at 580


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Recrystallization 4s at 580


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Complete recrystallization 8s at 580


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Grain Growth at 15 minutes at 580


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Grain Growth at 10 minutes at 700


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Recrystallization temperature

Temperature at whichrecrystallization reaches completion

at 1 hour.

Brass at 580oC complete

recrystallization at 8secRecrystallization temperature for

brass is 450 oC


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iron


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Grain Growth

Ktdd nn 0

N=2


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Ktdd nn 0

The average grain diameter for some metal alloy was measured as


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Time Grain diameter(min) (mm)

30 0.04991 0.071

The average grain diameter for some metal alloy was measured asfunction of time at 650C, which is tabulated below at two differenttimes:Give the equation.Calculate KWhat was the original grain diameter?What grain diameter would you predict after135min at 650C?

K=4/73E-5mm2/min

d0=.01mm

D135=7.98mm


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****Review for MSE 2034 Test 1

****All Quizzes and Homeworks********

Definitions and formulae:

Chapter 2 Fig 2.8; ionic bond, covalent bond, metallic bond, Van

der Waal Bond

Chapter 3

crystal structure, unit cell (repeating unit undertranslation), simple cubic, face centered cubic,coordination number, body centered cubic, hexagonalclosest packing, atomic packing factor, grain, lattice,diffraction, nl=2d sinq, diffraction angle = 2q

Chapter 4

vacancy, NV = N exp(-QV/kT), alloy, substitutionalpoint defect, interstitial, point defect, edge dislocation(Burgers vector perpendicular to dislocation line),screw dislocation (Burgers vector parallel todislocation line), twin boundary


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Review for MSE 2034 Test 1

Chapter 5

diffusion, self-diffusion, inter diffusion or impuritydiffusion, vacancy diffusion, interstitial diffusion,diffusion flux J = M/At, Ficks first law J = -D(dC/dx), D= D0 exp(-Qd/RT)

Chapter 6

stress, strain, Youngs modulus (modulus ofelasticity), Poissons ratio, yield strength, tensilestrength, ductility, resilience, toughness, hardness

Chapter 7

slip system, slip plane, slip direction, critical resolvedshear stress, lattice strain, strengthening by grain sizereduction, solid solution hardening and strainhardening, recovery, recrystalization, Recrystalizationtemperature, grain growth, R= cosq cosl ; y = 0 +

kd-1/2

, dn

- d0n

= k t

Relate Yield strength to three


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Relate Yield strength to three

strengthening processes

MSE 2034 Dislocations and Strengthening Mechanisms

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