chapter 7: dislocations & strengthening...

ISSUES TO ADDRESS:

• Why are dislocations observed primarily in metals

and alloys?

• How are strength and dislocation motion related?

• How do we increase strength?

• How can heat treatment change strength and other properties?

Chapter 7:

Dislocations & Strengthening Mechanisms

Reference: Callister chapter 7 and 8th ed.

Dislocation Motion

Dislocations & plastic deformation

• Cubic & hexagonal metals - plastic deformation is 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 8e.

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 (relatively close-packed)

in <110> directions (close-packed)

=> total of 12 slip systems in FCC

– in BCC & HCP other slip systems occur

Deformation Mechanisms Adapted from Fig. 7.6, Callister 8e.

Slip Systems in Cubic Metals:

Note: By definition is the angle between

the stress direction and Slip direction; is

the angle between the normal to slip

plane and stress direction

Stress and Dislocation Motion

• Crystals slip due to a resolved shear stress, tR. • Applied tension can produce such a stress.

slip plane

normal, ns

Resolved shear stress: tR = F s /A s

AS

tR

tR

FS

Relation between s and tR

tR = FS /AS

F cos A / cos

F

FS

nS

AS A

Applied tensile stress: = F/A s

F A

F

st coscosR

Adapted from 7.5, Callister 8e.

• Condition for dislocation motion: CRSS ttR

• Crystal orientation can make

it easy or hard to move dislocation 10-4 GPa to 10-2 GPa

typically

st coscosR

Critical Resolved Shear Stress

tR = 0

=90°

s

tR = s /2 =45° =45°

s

tR = 0

=90°

s

Generally:

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

And

tCRSS = sy /2 for dislocations to move (in single crystals)

Determining and angles for Slip in Crystals

(single crystals this is easy!)

• and angles are respectively angle between tensile direction and Normal to Slip plane and angle between tensile direction and slip direction (these slip directions are material dependent)

• In General for cubic xtals, angles between directions are given by:

• Thus for metals we compare Slip System (normal to slip plane is a direction with exact indices as plane) to applied tensile direction using this equation to determine the values of and to plug into the tR equation to determine if slip is expected

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

• Stronger since grain boundaries

pin deformations

• Slip planes & directions

(, ) change from one

crystal to another.

• tR will vary from one

crystal to another.

• The crystal with the

largest tR yields first.

• Other (less favorably

oriented) crystals

yield (slip) later.

Slip Motion in Polycrystals s

300 mm


After seeing the effect of poly crystalline materials

we can say (as related to strength):

• Ordinarily ductility is sacrificed when an alloy is strengthened.

• The relationship between dislocation motion and mechanical behavior of metals is significance to the understanding of strengthening mechanisms.

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

• Virtually all strengthening techniques rely on this simple principle: Restricting or Hindering dislocation motion renders a material harder and stronger.

• We will consider strengthening single phase metals by: grain size reduction, solid-solution alloying, and strain hardening

Strategies for Strengthening:

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: 21 /

yoyield dk ss

Hall-Petch equation:

Adapted from Fig. 715., Callister 8e.

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 small crystallographic

misalignment across the boundary.

Boundaries between two different phases are also impediments to

movements of dislocations.

Impurity atoms distort the lattice & generate stress.

Stress can produce a barrier to dislocation motion.


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

Stress Concentration at Dislocations


Strengthening by Alloying

• small impurities tend to concentrate at dislocations on the

“Compressive stress side”

• reduce mobility of dislocation → increase strength

Strengthening by alloying

• Large impurities concentrate at dislocations on

“Tensile Stress” side – pinning dislocation

Example: Solid Solution Strengthening in Copper

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

• Empirical relation:

• Alloying increases sy and TS.

21 /

y C~sAdapted from Fig.

7.16 (a) and (b),

Callister 8e.

Tensile

str

ength

(M

Pa)

wt.% Ni, (Concentration C)

200

300

400

0 10 20 30 40 50 Yie

ld s

trength

(M

Pa)

wt.%Ni, (Concentration C)

60

120

180

0 10 20 30 40 50

• Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC

in Iron or Aluminum).

• Result:

sGb

y


3. Precipitation Strengthening

Slipped part of slip plane

Side View

precipitate

Top View Unslipped part of slip plane

λ

Large shear stress needed to move

dislocation toward precipitate and

shear it.

Dislocation “advances” but

precipitates act as “pinning” sites

with spacing λ. which “multiplies”

Dislocation density


4. Cold Work (%CW)

• Room temperature deformation.

• Common forming operations change the cross

sectional area:

Adapted from Fig.

11.8, Callister 7e.

-Forging

A o A d

force

die

blank

force -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

• Ti alloy after cold working:

• Dislocations entangle and

multiply

• Thus, Dislocation motion

becomes more difficult.

Adapted from Fig.

4.6, Callister 8e.

(Fig. 4.6 is courtesy

of M.R. Plichta,

Michigan

Technological

University.)

During Cold Work

0.9 mm

Result of Cold Work

Dislocation density (rd) =

– 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 rd increases:

total dislocation length

unit volume

large hardening

small hardening

s

e

s y0

s y1

Impact of Cold Work, low Carbon steel:

Lo-Carbon Steel!

Adapted from Fig. 7.20,

Callister 8e.

• Yield strength (sy) increases.

• Tensile strength (TS) increases.

• Ductility (%EL or %AR) decreases.

As cold work is increased

• What is the tensile strength &

ductility after cold working?

%6.35100 x %2

22

o

do

r

rrCW

Cold Work Analysis

Copper

D0=15,2 mm Dd=12,2 mm

Cold worked rod

• What is the tensile strength &

ductility after cold working to 35,6%?

Adapted from Fig. 7.19, Callister 8e. (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.)

Cold Work Analysis

% 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

• 1 hour treatment at Tannealing

decreases TS and increases %EL.

• Effects of cold work are reversed!

• 3 Annealing

stages to

discuss...

Effect of Heating After %CW tensile

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

• New grains are formed that:

-- have a low dislocation density

-- are small

-- consume cold-worked grains

Adapted from

Fig. 7.21 (a),(b),

Callister 8e.

(Fig. 7.21 (a),(b)

are courtesy of

J.E. Burke,

General Electric

Company.)

33% cold

worked

brass

New crystals

nucleate after

3 sec. at 580C.

0.6 mm 0.6 mm

Recrystallization of Brass

• All cold-worked grains are consumed = recrystallization,

example heat treatment at 580 °C

Adapted from

Fig. 7.21 (c),(d),

Callister 8e.

(Fig. 7.21 (c),(d)

are courtesy of

J.E. Burke,

General Electric

Company.)

After 4

seconds

After 8

seconds

0.6 mm 0.6 mm

Further Recrystallization of Brass

Partial recrystallization Complete recrystallization

• 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 8e.

(Fig. 7.21 (d),(e)

are courtesy of

J.E. Burke,

General Electric

Company.)

Grain Growth of Brass

• Empirical Relation:

Ktdd n

o

n

coefficient dependent on

material and temperature

Grain diameter at time t:

elapsed time exponent typical ~ 2

This is: Ostwald Ripening

TR

Adapted from Fig.

7.22, Callister 8e.

º

º

TR = recrystallization

temperature

Recrystallization Temperature, TR

TR = recrystallization temperature = point of highest rate

of property change

1. TR 0.3-0.6 Tm (K)

2. Due to diffusion annealing time TR = f(t) shorter

annealing time => higher TR

3. Higher %CW => lower TR – strain hardening

4. Pure metals lower TR due to dislocation movements

• Easier to move in pure metals => lower TR

Coldwork Calculations

A cylindrical rod of brass originally 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 380 MPa and a

ductility of at least 15 %EL are desired. Further more,

the final diameter must be 7,6 mm.

Explain how this may be accomplished.

Coldwork Calculations Solution

a) If we directly draw to the final diameter what happens?

%5,44100 x 2,10

6,71100 x

4

41

100 1100 x %

2

2

2

o

f

o

f

o

fo

D

D

xA

A

A

AACW

D o = 10,2 mm

Brass

Cold Work

D f = 7,6 mm

Coldwork Calculation Solution:

• For %CW = 44,5%

540 420

– sy = 420 MPa

– TS = 540 MPa > 380 MPa

6

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

From Callister figure 7.19 – ed. 8

Coldwork Calculation Solution:

380

12

15

27

For %EL > 15

For TS > 380 MPa > 12 %CW

< 27 %CW

b) our working range is limited to %CW = 12 – 27%

From Callister figure 7.19 – ed. 8

This process Needs an Intermediate Recrystallization

i.e.: 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

Summary:

1. Cold work D01= 10,2 mm Df1 = 8,5 mm

2. Anneal above Ds2 = Df1

3. Cold work Ds2= 8,5 mm Df 2 = 7,6 mm

Therefore, meets all requirements.

b) Coldwork Calculations Solution

20100 5,8

6,71%

2

2

xCW

24%

MPa 410

MPa 340

EL

TS

ys

Seen from Figure 7.19, ch. 7 Callister

• 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: dislocations & strengthening...

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