Chapter 7: Dislocations & Strengthening Mechanismssite.iugaza.edu.ps/mhaiba/files/2010/02/CH7... · WHY STUDY Dislocations and Strengthening Mechanisms? Strengthening mechanisms will

Chapter 7 Dislocations & Strengthening Mechanisms

11/23/2013

5:41 AM

Dr. Mohammad Suliman Abuhaiba, PE 1

WHY STUDY Dislocations and Strengthening Mechanisms?

Understand underlying mechanisms

of the techniques used to

strengthen & harden metals and

their alloys

Design & tailor mechanical

properties of materials

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Dr. Mohammad Suliman Abuhaiba, PE

2

WHY STUDY Dislocations and Strengthening Mechanisms?

Strengthening mechanisms will be

applied to the development of

mechanical properties for steel alloys

Why heat treating a deformed metal

alloy makes it softer & more ductile

and produces changes in its

microstructure

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1. Describe edge & screw dislocation motion

2. Describe how plastic deformation occurs by

motion of edge & screw dislocations in response

to applied shear stresses

3. Define slip system

4. Describe how grain structure of a polycrystalline

metal is altered when it is plastically deformed.

5. Explain how grain boundaries impede dislocation

motion & why a metal having small grains is

stronger than one having large grains.

Learning Objectives

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5. Describe & explain solid-solution strengthening

for substitutional impurity atoms in terms of lattice

strain interactions with dislocations

6. Describe & explain phenomenon of strain

hardening in terms of dislocations and strain field

interactions.

7. Describe recrystallization in terms of both

alteration of microstructure & mechanical

characteristics of the material

8. Describe phenomenon of grain growth from both

macroscopic and atomic perspectives.

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Learning Objectives

7.2 BASIC CONCEPTS

Dislocation Motion Cubic & hexagonal metals - plastic deformation is

by plastic shear or slip where one plane of atoms

slides over adjacent plane by dislocations motion.

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7.3 Characteristics of Dislocations

When metals are plastically

deformed:

Around 5% of deformation energy is

retained internally

remainder is dissipated as heat

The major portion of stored

energy is strain energy associated

with dislocations

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Figure 7.4: some atomic lattice distortion exists

around dislocation line because of presence of

extra half-plane of atoms.


Atoms immediately above and

adjacent to dislocation line are

squeezed together.

These atoms may be thought of as

experiencing a compressive strain relative

to atoms positioned in the perfect crystal

and far removed from the dislocation

Directly below half-plane, lattice

atoms sustain an imposed tensile strain

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Shear strains also exist in the vicinity of

edge dislocation

For a screw dislocation, lattice strains are

pure shear only

These lattice distortions may be

considered to be strain fields that radiate

from the dislocation line

The strains extend into the surrounding

atoms, and their magnitude decreases

with radial distance from the dislocation.

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7.3 Characteristics of Dislocations Strain fields surrounding

Figure 7.5 (a): Two edge dislocations of same sign

& lying on same slip plane exert a repulsive force

on each other

Compressive & tensile strain fields for both lie on

same side of the slip plane; there exists between

these two isolated dislocations a mutual repulsive

force that tends to move them apart

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Figure 7.5 (b) Edge dislocations of opposite

sign & lying on same slip plane exert an

attractive force on each other. Upon

meeting, they annihilate each other and

leave a region of perfect crystal

7.3 Characteristics of Dislocations strain fields surrounding

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7.4 Slip Systems Slip plane - plane allowing easiest slippage

Wide inter-planar spacing - highest planar densities

Slip direction - direction of movement - Highest

linear densities

FCC Slip occurs on {111} planes in <110> directions

=> total of 12 slip systems in FCC

in BCC & HCP other slip systems occur

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Table 7.1: Slip Systems for FCC,

BCC, and HCP Metals

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7.5 Slip in Single Crystals

coscosR

Crystals slip due to a

resolved shear stress, R.

Applied tension can produce

such a stress.

= angle between stress

direction & Slip direction

= angle between stress

direction & normal to slip

plane

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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

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Critical Resolved Shear Stress

R = 0

=90°

R = /2 =45° =45°

R = 0

=90°

7.5 Slip in Single Crystals

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Resolved (shear stress) is maximum at = = 45º

And

CRSS = y/2 for dislocations to move (in single crystals)

Determining & angles for Slip in

Crystals

For cubic crytals, angles between directions are given by:

1 1 2 1 2 1 2

2 2 2 2 2 2

1 1 1 2 2 2

u u v v w wCos

u v w u v w

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EXAMPLE PROBLEM 7.1 Resolved Shear Stress and Stress-to-Initiate-Yielding

Computations

Consider a single crystal of BCC iron oriented such

that a tensile stress is applied along a [010] direction.

a. Compute the resolved shear stress along a (110)

plane and in a direction when a tensile stress

of 52 MPa is applied.

b. If slip occurs on a (110) plane and in a [-111]

direction, and the critical resolved shear stress is

30MPa, calculate the magnitude of the applied

tensile stress necessary to initiate yielding.

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Stronger since grain boundaries

pin deformations

Slip planes & directions (, )

change from one crystal to

another

R will vary from one crystal to

another.

Crystal with largest R yields first

Other (less favorably oriented)

crystals yield (slip) later.

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7.6 Plastic Deformation of Polycrystalline Materials

300 mm


Ordinarily ductility is sacrificed when an alloy is strengthened.

Relationship between dislocation motion & mechanical behavior of metals is significance to understanding of strengthening mechanisms

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

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Strengthening principle: Restricting or Hindering dislocation motion renders a material harder & stronger.

We will consider strengthening single phase metals by:

1. Grain size reduction

2. Solid-solution alloying

3. Strain hardening

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7.8 Strengthening by Grain Size Reduction

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Grain boundaries are barriers to slip

Barrier "strength" increases with Increasing

angle of mis-orientation.

Smaller grain size: more barriers to slip

Hall-Petch equation:

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21 /

yoyield dk

7.8 Strengthening by Grain

Size Reduction Figure 7.15: Influence of

grain size on yield

strength of a 70 Cu–30 Zn

brass alloy

Increase Rate of solidification from the liquid

phase

Perform Plastic deformation followed by an

appropriate heat treatment

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7.8 Strengthening by Grain Size

Reduction Grain Size Reduction Techniques:

Grain size reduction improves toughness of many

alloys.

Small-angle grain boundaries are not effective in

interfering with the slip process because of small

crystallographic misalignment across the boundary.

Boundaries between two different phases are also

impediments to movements of dislocations.

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7.8 Strengthening by Grain Size Reduction

Impurity atoms distort the lattice & generate stress.

Stress can produce a barrier to dislocation motion.

7.9 Solid-solution Strengthening

• 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

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small impurities tend to concentrate at

dislocations on the “Compressive stress side”

reduce mobility of dislocation increase

strength

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Large impurities concentrate at dislocations on

“Tensile Stress” side – pinning dislocation


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

• Empirical relation:

• Alloying increases y and TS

21 /

y C~

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-Forging

A o A d

force

die blank

force

7.10 Strain Hardening Room temperature deformation

Common forming operations change cross

sectional area:

-Drawing

tensile force

A o A d die

die

-Extrusion

ram billet

container

container

force die holder

die

A o

A d extrusion

100 x %o

do

A

AACW

-Rolling

roll

A o A d

roll

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Ti alloy after cold working:

Dislocations entangle &

multiply

Thus, Dislocation motion

becomes more difficult

0.9 mm

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7.10 Strain Hardening

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 length

unit volume

large hardening

small hardening

e

y0

y1

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Figure 7.20:

influence of cold

work on stress

strain behavior of a

low-carbon steel

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What is the tensile strength & ductility after cold

working?

%6.35100 x %2

22

o

do

r

rrCW

Cold Work Analysis

D o =15.2mm

Cold

Work

D d =12.2mm

Copper

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What is the tensile strength & ductility after cold working to

35.6%?

% Cold Work

100

300

500

700

Cu

20 0 40 60

yield strength (MPa)

% Cold Work

tensile strength (MPa)

200

Cu

0

400

600

800

20 40 60

340MPa

TS = 340MPa

ductility (%EL)

% Cold Work

20

40

60

20 40 60 0 0

Cu

7%

%EL = 7% YS = 300 MPa

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Cold Work Analysis

Results for polycrystalline iron:

y & TS decrease with increasing temperature

%EL increases with increasing temperature

- e Behavior vs. Temperature

-200C

-100C

25C

800

600

400

200

0

Strain

Str

ess (

MP

a)

0 0.1 0.2 0.3 0.4 0.5

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Recovery, Recrystallization,

and Grain Growth

Properties & structures may revert back to

the precold-worked states by appropriate

heat treatment (annealing treatment).

Such restoration results from different

processes that occur at elevated

temperatures:

1. Recovery

2. Recrystallization

3. Grain growth

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1 hour treatment at

Tanneal... decreases TS and

increases %EL.

Effects of cold work

are reversed!

3 Annealing stages

to discuss...

Recovery, Recrystallization, & Grain

Growth

te

nsile

str

en

gth

(M

Pa

)

du

ctilit

y (

%E

L) tensile strength

ductility

600

300

400

500

60

50

40

30

20

annealing temperature (ºC) 200 100 300 400 500 600 700

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7.11 RECOVERY During recovery, some of the stored internal strain

energy is relieved by virtue of dislocation motion,

due to enhanced atomic diffusion at elevated

temperature.

There is some reduction in number of dislocations,

and dislocation configurations are produced

having low strain energies.

Physical properties such as electrical and thermal

conductivities are recovered to their precold-

worked states.

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Even after recovery is complete, the grains are still in a

relatively high strain energy state.

Recrystallization = formation of a new set of strain-

free and equiaxed grains that have low dislocation

densities and are characteristic of the pre-cold-worked

condition.

The driving force to produce this new grain structure is

the difference in internal energy between the strained

and unstrained material.

The new grains form as very small nuclei and grow

until they completely consume the parent material,

processes that involve short-range diffusion.

7.12 Recrystallization

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33% cold

worked

brass

New crystals

nucleate after

3 sec. at 580C.

0.6 mm 0.6 mm


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Figure 7.21

All cold-worked grains are consumed.

After 4

seconds

After 8

seconds

0.6 mm 0.6 mm

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Recrystallization temperature =

temperature at which recrystallization just

reaches completion in 1 h.

Recrystallization temperature for brass alloy of Figure 7.22 is about 450°C.

Typically, between one-third and one-half

of absolute melting temperature of a

metal or alloy

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TR

º

º

TR = recrystallization

temperature

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Figure 7.22: Annealing

temperature influence

(for an annealing time

of 1 h) on tensile

strength & ductility of a

brass alloy.


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Figure 7.23

Critical degree of

cold work below

which

recrystallization

cannot be made

to occur (between

2% and 20% cold

work)

Recrystallization proceeds more

rapidly in pure metals than in

alloys.

During recrystallization, grain

boundary motion occurs as the

new grain nuclei form and then

grow.

For pure metals, TRC = 0.4Tm

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At longer times, larger

grains consume

smaller ones.

Grain boundary area

(and therefore energy)

is reduced.

Empirical Relation:

After 8 s,

580ºC

After 15 min,

580ºC

0.6 mm 0.6 mm

7.13 Grain Growth

Ktdd n

o

n

coefficient dependent on

material & Temp.

grain dia. At time t.

elapsed time exponent typ. ~ 2

This is: Ostwald Ripening

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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.

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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

D o = 0.40 in

Brass

Cold Work

D f = 0.30 in

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Coldwork Calc Solution: Cont.

For %CW = 43.8%

540 420

– y = 420 MPa

– TS = 540 MPa > 380 MPa

6

– %EL = 6 < 15

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

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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%

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This process Needs an Intermediate

Recrystallization

Cold 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:

2 2

2 2

2 2

2 2

%% 1 100 1

100

f f

s s

D D CWCW x

D D

0.5

2

2

%1

100

f

s

D CW

D

2

2 0.5%

1100

f

s

DD

CW

0.5

1 2

200.30 1 0.335 in

100f sD D

Intermediate diameter =

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Coldwork Calculations Solution Summary:

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

2. Anneal above Ds2 = Df1

3. Cold work Ds2= 0.335 in Df 2 =0.30 in

Therefore, meets all requirements

20100 3350

301%

2

2

x

.

.CW 24%

MPa 400

MPa 340

EL

TS

y

%CW1 10.335

0.4

2

x 100 30

Fig 7.19

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Design Example 7.1

Description of Diameter Reduction Procedure

A cylindrical rod of non-cold-worked brass having an

initial diameter of 6.4 mm is to be cold worked by

drawing such that the cross-sectional area is

reduced. It is required to have a cold-worked yield

strength of at least 345 MPa and a ductility in excess

of 20 %EL; in addition, a final diameter of 5.1 mm is

necessary. Describe the manner in which this

procedure may be carried out.

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Dislocations are observed primarily in metals and

alloys.

Strength is increased by making dislocation motion

difficult.

Particular ways to increase strength are to:

1. decrease grain size

2. solid solution strengthening

3. precipitate strengthening

4. cold work

Heating (annealing) can reduce dislocation density

and increase grain size. This decreases the strength

.

Summary

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HomeWork Assignment

7, 12, 17, 24, 29, 41, D6

Due Tuesday 26/11/2013

Quiz is on Tuesday 3/12/2013

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Documents

Chapter 7: Dislocations & Strengthening Mechanismssite.iugaza.edu.ps/mhaiba/files/2010/02/CH7... · WHY STUDY Dislocations and Strengthening Mechanisms? Strengthening mechanisms will