strengthening mechanisms - of Bahanajar [PDF]

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 ???

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Strategies for Strengthening Metals

1. Reduce Grain Size • Grain boundaries are barriers to slip. • Barrier "strength" increases with Increasing angle of misorientation. • Smaller grain size: more barriers to slip.

• Hall-Petch Equation: 24/10/2010

Adapted from Fig. 8.14, Callister & Rethwisch 3e. (Fig. 8.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

 yield   o  k y d  1 / 2 Anrinal - ITP

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Grain Size Reduction Techniques: Increase Rate of solidification from the liquid phase. Perform Plastic deformation followed by an appropriate heat treatment. Notes:  Grain size reduction also improves toughness of many alloys. Small-angle grain boundaries are not effective in interfering with the slip process because of the slight crystallographic misalignment across the boundary. Boundaries between two different phases are also impediments to movements of dislocations. 24/10/2010

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GRAIN SIZE STRENGTHENING: AN EXAMPLE • 70wt%Cu-30wt%Zn brass alloy

grain size, d (mm)

 yield  o  k y d 1/ 2

20 0

• Data:

10 -1

10 -2 5x10 -3

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M(

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.)

10 0

d l ) eai P y

150

50 0 0

ky 1 4

8

12

[grain size (mm)]

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-0.5 8

Strategies for Strengthening Metals: 2. Solid Solutions • Impurity atoms distort the lattice & generate stress. • Stress can produce a barrier to dislocation motion. • Smaller substitutional impurity

• Larger substitutional impurity

A

C D

B

Impurity generates local stress at A and B that opposes dislocation motion to the right. 24/10/2010

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

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• 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.

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Interactions of the Stress Fields

COMPRESSIVE TENSILE

TENSILE 24/10/2010

COMPRESSIVE Anrinal - ITP

Interactions of the Stress Fields

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Stress Concentration at Dislocations

Adapted from Fig. 8.4, Callister & Rethwisch 3e.

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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 !

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Strengthening by Alloying • small impurities tend to concentrate at dislocations • reduce mobility of dislocation  increase strength

Adapted from Fig. 7.17, Callister 7e.

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Strengthening by alloying • large impurities concentrate at dislocations on low density side

Adapted from Fig. 7.18, Callister 7e.

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Ex: Solid Solution Strengthening in Copper • Tensile strength & yield strength increase with wt% Ni. 180 Adapted from Fig. 7.16 (a) and (b), Callister 7e.

400 120

300

60

0 10 20 30 40 50

wt.% Ni, (Concentration C)

• Empirical relation:

) aPM ( ht gnert s dl ei Y

) aPM ( ht gnert s eli sneT

200

wt.%Ni, (Concentration C)

y ~ C1/ 2

• Alloying increases y and TS. 24/10/2010

0 10 20 30 40 50

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3: Precipitation Strengthening • Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). precipitate Large shear stress needed to move dislocation toward precipitate and shear it.

Side View

Top View

Unslipped part of slip plane

S Slipped part of slip plane

• Result: 24/10/2010

1 y ~ S Anrinal - ITP

Dislocation “advances” but precipitates act as “pinning” sites with spacing S.

Application: Precipitation Strengthening • Internal wing structure on Boeing 767 Adapted from chapteropening photograph, Chapter 11, Callister 5e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.)

• 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.)

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1.5m

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4: Cold Work (%CW) • Room temperature deformation. • Common forming operations change the cross sectional area: -Forging

force

die A o blank

-Drawing die Ao

-Rolling Ad

Ao Adapted from Fig. 11.8, Callister 7e.

-Extrusion Ao tensile force

force

die

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Ad

roll

force Ad

roll

container

ram

billet

container

Ao  Ad %CW  x 100 Anrinal Ao - ITP

die holder extrusion

die

Ad

Effects of Stress at Dislocations Adapted from Fig. 8.5, Callister & Rethwisch 3e.

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Impact of Cold Work As cold work is increased • Yield strength (y) increases. • Tensile strength (TS) increases. • Ductility (%EL or %AR) decreases.

Adapted from Fig. 8.20, Callister & Rethwisch 3e.

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Cold Work Analysis • What is the tensile strength & ductility after cold working? ro2  rd2 %CW  x 100  35.6% 2 ro yield strength (MPa)

800

500

600 300MPa

100 0

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Cu

Do =15.2mm

tensile strength (MPa)

700

300

Copper Cold Work

40

% Cold Work y = 300MPa

60

40 20

400 340MPa 200 0

ductility (%EL)

60

Cu 20

Dd =12.2mm

20

40

60

% Cold Work TS = 340MPa

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Cu

7%

00

20

40

60

% Cold Work %EL = 7%

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-  Behavior vs. Temperature • Results for polycrystalline iron: 800

-200C

600

-100C

) a P M( s s ert S

400

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25C

200 0 0

0.1

0.2

0.3

0.4

0.5

Strain Adapted from Fig. 7.14, Callister & Rethwisch 3e.

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Coldwork Calculations Solution If we directly draw to the final diameter what happens? Brass Cold Work

Do = 0.40 in

Df = 0.30 in

A  A   A  o f %CW    x 100  1 f  x 100  Ao   Ao  2     D2 4   0.30   1 f2  x 100  1  x 100  43.8%   0.40    Do 4    24/10/2010

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Coldwork Calc Solution: Cont.

420

540

6

•

For %CW = 43.8% – y = 420 MPa – TS = 540 MPa – %EL = 6 < 15

Adapted from Fig. 8.19, Callister & Rethwisch 3e.

> 380 MPa

• This doesn’t satisfy criteria…… what can we do? 24/10/2010

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Coldwork Calc Solution: Cont.

15

380

27

12

For TS > 380 MPa

> 12 %CW

For %EL > 15

< 27 %CW

Adapted from Fig. 8.19, Callister & Rethwisch 3e.

 our working range is limited to %CW = 12-27 24/10/2010

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