1. Strengthening and recrystallization of plastically deformed metals. 2. Material selection according to the mechanical properties

Lecture 6

Why study strengthening mechanisms?

We can tailor the mechanical properties of an engineering material by choosing between strength and toughness

Plastic Deformation Plastic deformation is permanent, and

strength and hardness are measures of a material’s resistance to this deformation.

On a microscopic scale, plastic deformation corresponds to the net movement of large numbers of atoms in response to an applied stress.

In crystalline solids, plastic deformation most often involves the motion of dislocations

The ability of a metal to plastically deform depends on the ability of dislocations to move

4Dislocation Motion 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!

5

Dislocation Motion The process by which plastic deformation is

produced by dislocation motion is termed slip Dislocation moves along slip plane in slip

direction perpendicular to dislocation line

Edge dislocation

Screw dislocation

Strengthening of Metals Good industrial alloys -> high strengths yet

some ductility and toughness Since hardness and strength are related to

the ease with which plastic deformation can be made to occur, by reducing the mobility of dislocations, the mechanical strength may be enhanced

In contrast, the more unconstrained the dislocation motion, the greater is the facility with which a metal may deform, and the softer and weaker it becomes

Restricting or hindering dislocation motion renders a material harder and stronger

Strengthening Methods grain size reduction, solid-solution alloying, precipitation

hardening/strengthening strain hardening/strengthening

Strategies for Strengthening: 1: Reduce Grain Size

21 /yoyield dk

8

• Grain boundaries are barriers to slip.

• Barrier "strength" increases with Increasing angle of misorientation.

• Smaller grain size: more barriers to slip.

• Hall-Petch Equation:

d is the average grain diameter. σo and ky are constants for a particular material

Dependence of Yield Strength on Grain Size

The influence of grain size on the yield strength of a 70 Cu–30 Zn brass alloy

10

Strategies for Strengthening: 2: Solid Solution Strengthening

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

• Smaller substitutional impurity

A

B

• Larger substitutional impurity

C

D

Stress Concentration at Dislocations

11

SSS - Impurity Atoms

SSS – Effects of Impurity Atoms The resistance to slip is greater when

impurity atoms are present because the overall lattice strain must increase if a dislocation is moved away from them.

The same lattice strain interactions will exist between impurity atoms and dislocations that are in motion during plastic deformation.

Thus, a greater applied stress is necessary to first initiate and then continue plastic deformation for solid-solution alloys, as opposed to pure metals

SSS – Strength and Ductility

Variation with nickel content of (a) tensilestrength, (b) yield strength, and (c) ductility (%EL) for copper–nickel alloys, showing strengthening.

Strategies for Strengthening: 3: Precipitation Strengthening Precipitation strengthening, also called age

hardening, is a heat treatment technique used to increase the yield strength of malleable materials.

It relies on changes in solid solubility with temperature to produce fine particles of an impurity phase, which impede the movement of dislocations, or defects in a crystal's lattice.

Precipitation in solids can produce many different sizes of particles, which have different properties.

Alloys must be kept at elevated temperature for hours to allow precipitation to take place. This time delay is called aging

16

Strategies for Strengthening: 3: Precipitation Strengthening

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

• Result:S

~y1

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

17Application: Precipitation Strengthening

• Internal wing structure on Boeing 767

• Aluminum is strengthened with precipitates formed by alloying.

1.5m

Strategies for Strengthening: 4: Strain-Hardening

18

• 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

doA

AACW

-Rolling

roll

AoAd

roll

19

Dislocations During Cold Work• Ti alloy after cold working:

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

0.9 m

Result of Cold WorkDislocation density =

Carefully grown single crystal 103 mm-2

Deforming sample increases density 109-1010 mm-2

Heat treatment reduces density 105-106 mm-2

20

• Yield stress increases as dislocation density increases:

total dislocation length

unit volume

large hardeningsmall hardening

y0 y1

Impact of Cold Work21

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

As cold work is increased

22

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

%6.35100 x % 2

22

o

dor

rrCW

% Cold Work

100

300

500

700

Cu

200 40 60

yield strength (MPa)

y = 300MPa

300MPa

% Cold Work

tensile strength (MPa)

200

Cu

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%

23

- Behavior vs. Temperature

• 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.

2. vacancies replace atoms on the disl. half plane

3. disl. glides past obstacle

-200C

-100C

25C

800

600

400

200

0Strain

Stre

ss (M

Pa)

0 0.1 0.2 0.3 0.4 0.5

1. disl. trapped by obstacle

obstacle

24

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

• 3 Annealing stages to discuss...

tens

ile s

treng

th (M

Pa)

duct

ility

(%E

L)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

25

RecoveryAnnihilation reduces dislocation density.

• Scenario 1

Results 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

Recrystallization Even after recovery is complete, the grains

are still in a relatively high strain energy state Recrystallization is the formation of a new

set of strain-free and equiaxed grains (i.e., having approximately equal dimensions in all directions) that have low dislocation densities and are characteristic of the precold-worked condition.

The new grains form as very small nuclei and grow until they completely consume the parent material, processes that involve short-range diffusion

27Recrystallization• New grains are formed that: -- have a small dislocation density -- are small -- consume cold-worked grains.

33% coldworkedbrass

New crystalsnucleate after3 sec. at 580C.

0.6 mm 0.6 mm

28Further Recrystallization• All cold-worked grains are consumed.

After 4seconds

After 8seconds

0.6 mm0.6 mm

Recrystallization During recrystallization, the mechanical properties

that were changed as a result of cold working are restored to their precold-worked values; that is, the metal becomes softer, weaker, yet more ductile

Recrystallization is a process the extent of which depends on both time and temperature. The degree (or fraction) of recrystallization increases with time

For pure metals, the recrystallization temperature is normally 0.3Tm where Tm is the absolute melting temperature

Grain Growth After recrystallization is complete, the

strain-free grains will continue to grow if the metal specimen is left at the elevated temperature; this phenomenon is called grain growth

Grain growth does not need to be preceded by recovery and recrystallization; it may occur in all polycrystalline materials, metals and ceramics alike

31

Grain Growth• 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

• Empirical Relation:

Ktdd no

n elapsed time

coefficient dependenton material and T.

grain diam.at time t.

exponent typical. ~ 2

Ostwald Ripening

After 10 min,700ºC

32

TR

º

º

TR = recrystallization temperature

Time and Temperature Dependent Grain Growth

Recrystallization Temperature, TR

TR = recrystallization temperature = point of highest rate of property change1. Tm => TR 0.3-0.6 Tm (K)2. Due to diffusion annealing time TR =

f(t) shorter annealing time => higher TR

3. Pure metals lower TR due to dislocation movements Easier to move in pure metals => lower TR

34

Coldwork CalculationsA cylindrical rod of brass originally 0.40in (10.2mm) 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 55,000psi (380MPa) and a ductility of at least 15%EL are desired. Furthermore, the final diameter must be 0.30in (7.6mm). Explain how this may be accomplished.

35

Coldwork Calculations SolutionIf we directly draw to the final diameter what

happens?

36

%843100 x 4003001100 x

441

100 1100 x %

2

2

2

...

DD

xAA

AAACW

o

f

o

f

o

fo

Do = 0.40 in

BrassCold Work

Df = 0.30 in

Coldwork Calc Solution: Cont.

For %CW = 43.8%

37

540420

y = 420 MPa– TS = 540 MPa > 380 MPa

6

– %EL = 6 < 15

• This doesn’t satisfy criteria…… what can we do?

Coldwork Calc Solution: Cont.

38

380

12

15

27

For %EL < 15

For TS > 380 MPa > 12 %CW

< 27 %CW

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

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:

39

100%1 100 1% 2

02

22

202

22 CW

DDx

DDCW ff

50

02

2

100%1

.f CW

DD

50

202

100%1

.f

CWDD

in 335.010020130.0

5.0

021

DD f

Intermediate diameter =

Coldwork Calculations Solution

Summary:1. Cold work D01= 0.40 in Df1 = 0.335 in2. Anneal to remove all CW effects; D02 = Df1

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

Therefore, meets all requirements

40

%20100 335.0

3.01%2

2

xCW

24%MPa 400

MPa 340

ELTSy

%30100 4.0

335.01%2

1

xCW

Material Selection According to the Mechanical Properties

Material Selection – The Basics Getting the optimum balance of performance,

quality, and cost requires a careful combination of material and part design

The ideal product is one that will just meet all requirements.

Anything better will usually incur added cost through higher-grade materials, enhanced processing, or improved properties that may not be necessary.

Anything worse will likely cause product failure, dissatisfied customers, and the possibility of unemployment

Material Selection – The Basics

The interdependence between materials and their processing must also be recognized.

New processes frequently accompany new materials, and their implementation can often cut production costs and improve product quality.

A change in material may well require a change in the manufacturing process

Improper processing of a well-chosen material can definitely result in a defective product.

Materials and Manufacturing

• An engineering material may possess different properties depending upon its structure.

• Processing of that material can alter the structure, which in turn will alter the properties.

• Altered properties certainly alter performance.

PROCEDURES FOR MATERIAL SELECTIONEvery engineered item goes through a sequence of activities that includes:design material selection process selection production evaluation possible redesign or modification

Requirements for Material Selection

(1) shape or geometry considerations, (2) property requirements, (3) manufacturing concerns

1. GEOMETRIC CONSIDERATIONS

What is the relative size of the component? How complex is its shape? What are the surface-finish requirements?

Must all surfaces be finished? Could a minor change in part geometry

increase the ease of manufacture or improve the performance (fracture resistance, fatigue resistance, etc.) of the part?

2. Mechanical Properties How much static strength is required? If the part is accidentally overloaded, is it

permissible to have a sudden brittle fracture, or is plastic deformation and distortion a desirable precursor to failure?

How much can the material bend, stretch, twist, or compress under load and still function properly?

Are any impact loadings anticipated? If so, of what type, magnitude, and velocity?

2. Mechanical Properties Can you envision vibrations or cyclic

loadings? If so, of what type, magnitude, and frequency?

Is wear resistance desired? Where? How much? How deep?

Will all of the above requirements be needed over the entire range of operating temperature? If not, which properties are needed at the lowest extreme? At the highest extreme?

Environmental Considerations What are the lowest, highest, and normal

temperatures the product will see? Will temperature changes be cyclic? How fast will temperature changes occur?

What is the most severe environment that is anticipated as far as corrosion or deterioration of material properties is concerned?

What is the desired service lifetime for the product?

3. Manufacturing Concerns

How many of the components are to be produced? At what rate?

What is the desired level of quality compared to similar products on the market?

Has the design addressed the requirements that will facilitate ease of manufacture?

E MCH 213D

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