chapter 7- chapter 7: dislocations and strengthening in ch.6 –plastic (permanent) & elastic...

Chapter 7-

CHAPTER 7: DISLOCATIONS AND STRENGTHENING

• In Ch.6– Plastic (permanent) & Elastic (reversible)

• Yield Strength and hardness are a measure of materials resistance to deformation

– In microscopic scale what is going on ?• Net movement of large number of atoms in response to

an applied stress• Involves movement of dislocations, in Ch. 4

– How does strengthening happen ?– Why do some materials are stronger than others ?– How can one manipulate dislocations and their

motion ?

Chapter 7 -

Dislocations & 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

++

++

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

+++++++

Chapter 7 -

Dislocation Motion

Dislocations & plastic deformation• Cubic & hexagonal metals - plastic deformation by

plastic shear or slip where one plane of atoms slides over adjacent plane by defect motion (dislocations).

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

Adapted from Fig. 7.1, Callister 7e.

Chapter 7- 3

• Produces plastic deformation !• Bonds are incrementally broken and re-formed. Much less force is needed , why ?

Plasticallystretchedzincsinglecrystal.

• If dislocations don't move, plastic deformation doesn't happen!

Adapted from Fig. 7.1, Callister 6e. (Fig. 7.1 is adapted from A.G. Guy, Essentials of Materials Science, McGraw-Hill Book Company,New York, 1976. p. 153.)

Adapted from Fig. 7.9, Callister 6e. (Fig. 7.9 is from C.F. Elam, The Distortion of Metal Crystals, Oxford University Press, London, 1935.)


DISLOCATION MOTION

SEE THE MOVIE AGAIN !

How do we generate the dislocation motion ?

SLIP PLANE

Chapter 7-

DISLOCATION MOTION

t0t1t2t3

The motion of a single dislocation across the plane causesthe top half of the crystal to move (to slip) with respect tothe bottom half but we can not have to break all the bondsacross the middle plane simultaneously (which wouldrequire a very large force).

The slip plane (O0-O1)– the crystallographic plane of dislocationmotion.

O1 O0

Initial statedeformed apply force

Chapter 7-

STRESS and Dislocation Motion

NOT be parallel to the dislocation line !

Dislocation line

Chapter 7 -

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


Chapter 7-

STRESS AND DISLOCATION MOTION

Chapter 7-

STRESS field around dislocations

WHY is there a stress field ?Atoms try to relax by trying to achieve their positions for the case if there was not a dislocation in the vicinity !

Bonds are Stretched

Bonds are Compressed

Chapter 7-

Stress fields of dislocations interacting !The strain fields around dislocations interact with each other. Hence, they exert force on each other.

Edge dislocations, when they are in the same plane, they repel each other if they have the samesign (direction of the Burgers vector). WHY ?

They can attract and annihilate if they have opposite signs. PROVE !

Chapter 7-

Dislocation Density

• The number of dislocations in a material is expressed using the term dislocation density - the total dislocation length per unit volume or the # of dislocations intersecting a unit area. Units are mm / mm3, or just / mm

• Dislocation densities can vary from 103 mm-2 in carefully solidified metal crystals to 1010 mm-2 in heavily deformed metals.

Where do Dislocations come from, what are their sources ?• Most crystalline materials, especially metals, have dislocations

in their as-formed state, mainly as a result of stresses (mechanical, thermal...) associated with the manufacturing processes used.

• The number of dislocations increases dramatically during plastic deformation.

• Dislocations spawn from existing dislocations, grain boundaries and surfaces and other “defects” .

NICE SIMULATIONS => http://www.ims.uconn.edu/centers/simul/movie/

Chapter 7 -

Slip System– Slip plane - plane allowing easiest slippage

• Wide interplanar spacings - highest planar densities

– Slip direction - direction of movement - Highest linear densities

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

=> total of 12 slip systems in FCC– in BCC & HCP other slip systems occur

Deformation Mechanisms


Chapter 7-

SLIP SYSTEMS !!!

• Dislocations move with ease on certain crystallographic planes and along certain directions on these planes ! – The plane is called a slip plane– The direction is called a slip direction – Combination of the plane of slip and direction is a

slip system• The slip planes and directions are those of highest

packing density. – The distance between atoms is shorter than the

average…High number of coordination along the planes also important !

PLS SEE TABLE 7.1 in ur books , page 180 !!!

Chapter 7-

Concept Check

Page 181, in 7th edition !

Chapter 7 -

Stress and Dislocation Motion• Crystals slip due to a resolved shear stress, R. • Applied tension can produce such a stress.

slip plane

normal, ns

Resolved shear stress: R =Fs/As

slip

directi

on

AS

R

R

FS

slip

directi

on

Relation between and R

R =FS /AS

Fcos A/cos

F

FS

nS

AS

A

Applied tensile stress: = F/A

slip

directi

on

FA

F

coscosR +

Chapter 7 -

• Condition for dislocation motion: CRSS R

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

10-4 GPa to 10-2 GPa

typically

coscosR

Critical Resolved Shear Stress

maximum at = = 45º

R = 0

=90°

R = /2=45°=45°

R = 0

=90°

Chapter 7-

Resolving the Applied Stress on a SLIP PLANE !

• Shear Stress has to be resolved on to the slip planes as;– Shear Stress is needed for dislocations to move / slip– Dislocations can only move on slip planes, and these

planes are rarely on axis with the applied force !

We should resolve the force applied in a tensile test, F, on to the cross-sectional area A where the slip is going to take place;

Chapter 7-

Critical Resolved Shear Stress => Slip in Single X’tals

Macroscopically;Q: when do materials plastically deform / yield ?

Stress > YSQ: How does plastic deformation take place ?

Dislocation Motion ( Elastic deformation ??)Q: On what planes does dislocations move ?

Slip PlanesQ: At stress = YS, what would be the minimum resolved shear stress needed to act on dislocations to initiate dislocation motion, onset of yield/plastic deformation ?

HIMMM ! Lets think !!!

Chapter 7-

The Critical Resolved Shear StressThe minimum shear stress required to initiate slip is termed:the critical resolved shear stress

Chapter 7 -

Single Crystal Slip



Plasticallystretched

zincsingle

crystal.

A large number of dislocations are generated on slip planes, as they leave the system they form these “shear bands”

Chapter 7 -

Ex: Deformation of single crystal

So the applied stress of 6500 psi will not cause the crystal to yield.

cos cos 6500 psi

=35°

=60°

(6500 psi) (cos35)(cos60)

(6500 psi) (0.41)

2662 psi crss 3000 psi

crss = 3000 psi

a) Will the single crystal yield? b) If not, what stress is needed?

= 6500 psi


Chapter 7 -

Ex: Deformation of single crystal

psi 732541.0

psi 3000

coscoscrss

y

What stress is necessary (i.e., what is the yield stress, y)?

)41.0(cos cos psi 3000crss yy

psi 7325 y

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

Chapter 7 -

• Stronger - grain boundaries pin deformations

• Slip planes & directions (, ) change from one crystal to another.

• R will vary from one crystal to another.

• The crystal with the largest R yields first.

• Other (less favorably oriented) crystals yield later.

Adapted from Fig. 7.10, Callister 7e.(Fig. 7.10 is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].)

Slip Motion in Polycrystals

300 m

Chapter 7-

Plastic Deformation in Polycrystalline Materials

Chapter 7 -

• Can be induced by rolling a polycrystalline metal

- before rolling

235 m

- isotropic since grains are approx. spherical & randomly oriented.

- after rolling

- anisotropic since rolling affects grain orientation and shape.

rolling direction

Adapted from Fig. 7.11, Callister 7e. (Fig. 7.11 is from W.G. Moffatt, G.W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 140, John Wiley and Sons, New York, 1964.)

Anisotropy in y

Chapter 7 -

side view

1. Cylinder of Tantalum machined from a rolled plate:

rolli

ng d

irect

ion

2. Fire cylinder at a target.

• The noncircular end view shows anisotropic deformation of rolled material.

endview

3. Deformed cylinder

platethicknessdirection

Photos courtesyof G.T. Gray III,Los AlamosNational Labs. Used withpermission.

Anisotropy in Deformation

Chapter 7- 10

1. Cylinder of Tantalum machined from a rolled plate: side view

endview

• The noncircular end view shows: anisotropic deformation of rolled material.

2. Fire cylinder at a target.

3. Deformed cylinder

platethicknessdirection

Photos courtesyof G.T. Gray III,Los AlamosNational Labs. Used withpermission.

ANISOTROPY IN DEFORMATION

Chapter 7-

STRENGTHENING MECHANISMS

The ability of a metal to deform depends on the ability of dislocations to move with relative ease under loading conditions

Restricting dislocation motion will inevitably make the material stronger, need more force to induce same amount of deformation !

• Mechanisms of strengthening in single-phase metals:– grain-size reduction– solid-solution alloying– strain hardening

• Ordinarily, strengthening mechanisms reduces ductility, why ???

Chapter 7- 7

• Grain boundaries are barriers to slip.• Barrier "strength" increases with mis-orientation.

High-angle boundaries are better in blocking slip !

• Smaller grain size: more barriers to slip.

Grain size can be changed by processing !

grain boundary

slip plane

grain Agr

ain

B

yield o kyd 1/2

Adapted from Fig. 7.12, Callister 6e.(Fig. 7.12 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

4 STRATEGIES FOR STRENGTHENING: 1: REDUCE GRAIN SIZE

Average grain size

Hall-Petch Equation :

Chapter 7- 8

• 70wt%Cu-30wt%Zn brass alloy

yield o kyd 1/2

• Data:

Adapted from Fig. 7.13, Callister 6e.(Fig. 7.13 is adapted from H. Suzuki, "The Relation Between the Structure and Mechanical Properties of Metals", Vol. II, National Physical Laboratory Symposium No. 15, 1963, p. 524.)

[grain size (mm)]-0.5

yie

ld(M

Pa)

50

100

150

200

04 8 12 16

10-1 10-2 5x10-3grain size, d (mm)

1

ky

0

GRAIN SIZE STRENGTHENING: AN EXAMPLE

Chapter 7 -

• Impurity atoms distort the lattice & generate stress.• Stress can produce a barrier to dislocation motion.

4 Strategies for Strengthening: 2: Solid Solutions

• Smaller substitutional impurity

Impurity generates local stress at A and B that opposes dislocation motion to the right.

A

B

• Larger substitutional impurity

Impurity generates local stress at C and D that opposes dislocation motion to the right.

C

D

Chapter 7-

Strategy #2: Solid Solutions

• Alloyed metals are usually stronger than their pure base metals counter parts.

Why ? Interstitial or substitutional impurities in a solution cause lattice strain, aka distortions in the lattice

Then ? • Strain field around the impurities interact with dislocation strain

fields and impede dislocation motion.• Impurities tend to diffuse and segregate around the dislocation

core to find atomic sites more suited to their radii. This reduces the overall strain energy and “anchor” the dislocation.Motion of the dislocation core away from the impurities moves it to a region of lattice where the atomic strains are greater, where lattice strains due to dislocation is no longer

compensated by the impurity atoms.

Chapter 7-

Interactions of the Stress Fields

TENSILE COMPRESSIVE

COMPRESSIVE

TENSILE

Chapter 7-

Interactions of the Stress Fields

Chapter 7 -

Stress Concentration at Dislocations


Chapter 7-

Impurity Segregation

Impurities tend to segregate at energetically favorable areas around the dislocation core and partially decrease the overall stress field generated around the dislocation core.

However, when stress is applied more load is needed to move dislocations with impurity atoms segregated to its core !

Chapter 7 -

Strengthening by Alloying

• small impurities tend to concentrate at dislocations• reduce mobility of dislocation increase strength


Chapter 7 -

Strengthening by alloying

• large impurities concentrate at dislocations on low density side


Chapter 7 -

Ex: Solid SolutionStrengthening in Copper

• Tensile strength & yield strength increase with wt% Ni.

• Empirical relation:

• Alloying increases y and TS.

21 /y C~

Adapted from Fig. 7.16 (a) and (b), Callister 7e.

Ten

sile

str

engt

h (M

Pa)

wt.% Ni, (Concentration C)

200

300

400

0 10 20 30 40 50 Yie

ld s

tren

gth

(MP

a)wt.%Ni, (Concentration C)

60

120

180

0 10 20 30 40 50

Chapter 7 -

• Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum).

• Result:S

~y

1

4 Strategies for Strengthening: 3: Precipitation Strengthening

Large shear stress needed to move dislocation toward precipitate and shear it.

Dislocation “advances” but precipitates act as “pinning” sites with

S.spacing

Side View

precipitate

Top View

Slipped part of slip plane

Unslipped part of slip plane

Sspacing

Chapter 7 -

• Internal wing structure on Boeing 767

• Aluminum is strengthened with precipitates formed by alloying.

Adapted from Fig. 11.26, Callister 7e. (Fig. 11.26 is courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

1.5m

Application: Precipitation Strengthening

Adapted from chapter-opening photograph, Chapter 11, Callister 5e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

Chapter 7 -

4 Strategies for Strengthening: 4: Cold Work (%CW)

• Room temperature deformation.• Common forming operations change the cross sectional area:


-Forging

Ao Ad

force

dieblank

force-Drawing

tensile force

AoAddie

die

-Extrusion

ram billet

container

containerforce

die holder

die

Ao

Adextrusion

100 x %o

do

A

AACW

-Rolling

roll

AoAd

roll

Chapter 7 -

• Ti alloy after cold working:

• Dislocations entangle with one another during cold work.• Dislocation motion becomes more difficult.

Adapted from Fig. 4.6, Callister 7e. (Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.)

Dislocations During Cold Work

0.9 m

Chapter 7 -

Result of Cold Work

Dislocation density =

– Carefully grown single crystal

ca. 103 mm-2

– Deforming sample increases density

109-1010 mm-2

– Heat treatment reduces density

105-106 mm-2

• Yield stress increases as d increases:

total dislocation lengthunit volume

large hardening

small hardening

y0 y1

Chapter 7- 18

• Dislocation density (d) goes up: Carefully prepared sample: d ~ 103

mm/mm3

Heavily deformed sample: d ~ 1010

mm/mm3

• Ways of measuring dislocation density:

ORlength, l1

length, l2length, l3

Volume, V

l1 l2 l3V

d dN

A

Area, A

N dislocation pits (revealed by etching)

dislocation pit

• Yield stress increases as d increases:

large hardeningsmall hardening

y0 y1

Micrograph adapted from Fig. 7.0, Callister 6e. (Fig. 7.0 is courtesy of W.G. Johnson, General Electric Co.)

40m

RESULT OF COLD WORK

Chapter 7-

• An increase in y due to plastic deformation.BUT actually # of dislocations are

increasing !!!

22

• Curve fit to the stress-strain response:

large hardening

small hardening

unlo

ad

relo

ad

y 0

y 1

T C T n“true” stress (F/A) “true” strain: ln(L/Lo)

hardening exponent: n=0.15 (some steels) to n=0.5 (some copper)

STRENGTHENING STRATEGY 4: COLD WORK (%CW)

Chapter 7-

Stress fields of dislocations interacting !The strain fields around dislocations interact with each other. Hence, they exert force on each other.

Edge dislocations, when they are in the same plane, they repel each other if they have the samesign (direction of the Burgers vector). WHY ?

They can attract and annihilate if they have opposite signs. PROVE !

Chapter 7 -

Effects of Stress at Dislocations


Chapter 7-

• Yield strength ( ) increases.• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.

21

y

Adapted from Fig. 7.18, Callister 6e. (Fig. 7.18 is from Metals Handbook: Properties and Selection: Iron and Steels, Vol. 1, 9th ed., B. Bardes (Ed.), American Society for Metals, 1978, p. 221.)

IMPACT OF COLD WORK

Str

ess

% cold work Strain

Chapter 7 -

Impact of Cold Work


• Yield strength (y) increases.• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.

As cold work is increased

Chapter 7 -

• What is the tensile strength & ductility after cold working?

Adapted from Fig. 7.19, Callister 7e. (Fig. 7.19 is adapted from Metals Handbook: Properties and Selection: Iron and Steels, Vol. 1, 9th ed., B. Bardes (Ed.), American Society for Metals, 1978, p. 226; and Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th ed., H. Baker (Managing Ed.), American Society for Metals, 1979, p. 276 and 327.)

%6.35100 x %2

22

o

do

r

rrCW

Cold Work Analysis

% Cold Work

100

300

500

700

Cu

200 40 60

yield strength (MPa)

y = 300MPa

300MPa

% Cold Work

tensile strength (MPa)

200Cu

0

400

600

800

20 40 60

ductility (%EL)

% Cold Work

20

40

60

20 40 6000

Cu

Do =15.2mm

Cold Work

Dd =12.2mm

Copper

340MPa

TS = 340MPa

7%

%EL = 7%

Chapter 7 -

• Results for polycrystalline iron:

• y and TS decrease with increasing test temperature.• %EL increases with increasing test temperature.• Why? Vacancies help dislocations move past obstacles.


- Behavior vs. Temperature

2. vacancies replace atoms on the disl. half plane

3. disl. glides past obstacle

-200C

-100C

25C

800

600

400

200

0

Strain

Str

ess

(M

Pa)

0 0.1 0.2 0.3 0.4 0.5

1. disl. trapped by obstacle

obstacle

Chapter 7 -

• 1 hour treatment at Tanneal... decreases TS and increases %EL.• Effects of cold work are reversed!

• 3 Annealing stages to discuss...

Adapted from Fig. 7.22, Callister 7e. (Fig.7.22 is adapted from G. Sachs and K.R. van Horn, Practical Metallurgy, Applied Metallurgy, and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, 1940, p. 139.)

Effect of Heating After %CW te

nsi

le s

tre

ngth

(M

Pa)

duc

tility

(%

EL

)tensile strength

ductility

Recovery

Recrystallization

Grain Growth

600

300

400

500

60

50

40

30

20

annealing temperature (ºC)200100 300 400 500 600 700

Chapter 7 -

Annihilation reduces dislocation density.

Recovery

• Scenario 1Results from diffusion

• Scenario 2

4. opposite dislocations

meet and annihilate

Dislocations annihilate and form a perfect atomic plane.

extra half-plane of atoms

extra half-plane of atoms

atoms diffuse to regions of tension

2. grey atoms leave by vacancy diffusion allowing disl. to “climb”

R

1. dislocation blocked; can’t move to the right

Obstacle dislocation

3. “Climbed” disl. can now move on new slip plane

Chapter 7 -

• New grains are formed that: -- have a small dislocation density -- are smalller then initial cold worked ones -- consume cold-worked grains.

Adapted from Fig. 7.21 (a),(b), Callister 7e. (Fig. 7.21 (a),(b) are courtesy of J.E. Burke, General Electric Company.)

33% coldworkedbrass

New crystalsnucleate after3 sec. at 580C.

0.6 mm 0.6 mm

Recrystallization

Chapter 7 -

• All cold-worked grains are consumed.

Adapted from Fig. 7.21 (c),(d), Callister 7e. (Fig. 7.21 (c),(d) are courtesy of J.E. Burke, General Electric Company.)

After 4seconds

After 8seconds

0.6 mm0.6 mm

Further Recrystallization

Chapter 7 -

• At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy) is reduced.

After 8 s,580ºC

After 15 min,580ºC

0.6 mm 0.6 mm

Adapted from Fig. 7.21 (d),(e), Callister 7e. (Fig. 7.21 (d),(e) are courtesy of J.E. Burke, General Electric Company.)

Grain Growth

• Empirical Relation:

Ktdd no

n elapsed time

coefficient dependenton material and T.

grain diam.at time t.

exponent typ. ~ 2

Ostwald Ripening

Chapter 7 -

TR


º

º

TR = recrystallization temperature

Chapter 7 -

Coldwork Calculations

A cylindrical rod of brass originally 0.40 in (10.2 mm) in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 55,000 psi (380 MPa) and a ductility of at least 15 %EL are desired. Further more, the final diameter must be 0.30 in (7.6 mm). Explain how this may be accomplished.

Chapter 7 -

Coldwork Calculations Solution

If we directly draw to the final diameter what happens?

%843100 x 400

3001100 x

4

41

100 1100 x %

2

2

2

..

.

D

D

xA

A

A

AACW

o

f

o

f

o

fo

Do = 0.40 in

BrassCold Work

Df = 0.30 in

Chapter 7 -

Coldwork Calc Solution: Cont.

• For %CW = 43.8%Adapted from Fig. 7.19, Callister 7e.

540420

y = 420 MPa– TS = 540 MPa > 380 MPa

6

– %EL = 6 < 15• This doesn’t satisfy criteria…… what can we do?

Chapter 7 -

Coldwork Calc Solution: Cont.


380

12

15

27

For %EL < 15

For TS > 380 MPa > 12 %CW

< 27 %CW

our working range is limited to %CW = 12-27

Chapter 7 -

Coldwork Calc Soln: RecrystallizationCold draw-anneal-cold draw again• For objective we need a cold work of %CW 12-27

– We’ll use %CW = 20• Diameter after first cold draw (before 2nd cold draw)?

– must be calculated as follows:

100

%1 100 1% 2

02

22

202

22 CW

D

Dx

D

DCW ff

50

02

2

100

%1

.

f CW

D

D

50

202

100%

1.

f

CW

DD

m 335010020

130050

021 ..DD.

f

Intermediate diameter =

Chapter 7 -

Coldwork Calculations Solution

Summary:

1. Cold work D01= 0.40 in Df1 = 0.335 m

2. Anneal above D02 = Df1

3. Cold work D02= 0.335 in Df 2 =0.30 m

Therefore, meets all requirements

20100 3350

301%

2

2

x

.

.CW

24%

MPa 400

MPa 340

EL

TSy

%CW1 10.335

0.4

2

x 100 30

Fig 7.19

Chapter 7 -

Rate of Recrystallization

• Hot work above TR

• Cold work below TR

• Smaller grains – stronger at low temperature– weaker at high temperature

t/RT

BCt

kT

ERtR

1:note

log

logloglog 0

RT1

log t

start

finish

50%

TR ~ 0.3 Tm-0.7 Tm

TR depends on %CW•Decrease with increasing % CW

No Strain hardening occurs

Chapter 7 -

• Dislocations are observed primarily in metals and alloys.

• Strength is increased by making dislocation motion difficult.

• Particular ways to increase strength are to: --decrease grain size --solid solution strengthening --precipitate strengthening --cold work

• Heating (annealing) can reduce dislocation density and increase grain size. This decreases the strength.

Summary

Chapter 7-

Reading: Chapter 7

Core Problems: 7.6, 7.7, 7.13 (see page 183-4), 7.14, 7.23, 7.31, 7.41 (figure 7.25)

Bonus Problems: 7.40, 7.36, 7.39

Due date is April 3th

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chapter 7- chapter 7: dislocations and strengthening in ch.6 –plastic (permanent) & elastic...

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