Chapter 7: Dislocations & Strengthening Mechanisms 9/15/2015 11:21 PM Dr. Mohammad Abuhaiba, PE1

Chapter 7:Dislocations & Strengthening Mechanisms

04/19/23 15:57

Dr. Mohammad Abuhaiba, PE1

WHY STUDY Dislocations and Strengthening Mechanisms?Understand underlying mechanisms

of the techniques used to strengthen and harden metals and their alloys.

Design and tailor mechanical properties of materials.

04/19/23 15:57

Dr. Mohammad 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 and more ductile and produces changes in its microstructure.

04/19/23 15:57


3

1. Describe edge and 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 and why a metal having small grains is stronger than one having large grains.

Learning Objectives

04/19/23 15:57


4

5. Describe & explain solid-solution strengthening for substitutional impurity atoms in terms of lattice strain interactions with dislocations.

6. Describe and explain phenomenon of strain hardening in terms of dislocations and strain field interactions.

7. Describe recrystallization in terms of both alteration of microstructure and mechanical characteristics of the material.

8. Describe phenomenon of grain growth from both macroscopic and atomic perspectives.

Learning Objectives

04/19/23 15:57


5

7.2 BASIC CONCEPTSDislocation Motion

Cubic & hexagonal metals - plastic deformation is by plastic shear or slip where one plane of atoms slides over adjacent plane by dislocations motion.

04/19/23 15:57


6

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.

04/19/23 15:57


7


04/19/23 15:57


8

Figure 7.4: some atomic lattice distortion exists around the dislocation line because of presence of extra half-plane of atoms.


Atoms immediately above and adjacent to dislocation line are squeezed together. As a result, 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 the half-plane, the lattice atoms sustain an imposed tensile strain.

04/19/23 15:57


9


Shear strains also exist in the vicinity of the 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.

04/19/23 15:57


10

7.3 Characteristics of Dislocationsstrain fields surrounding

Consider two edge dislocations that have the same sign and the identical slip plane (Figure 7.5a).

Compressive and tensile strain fields for both lie on the same side of the slip plane; there exists between these two isolated dislocations a mutual repulsive force that tends to move them apart.

04/19/23 15:57


11


Two dislocations of opposite sign and having same slip plane will be attracted to one another (Figure 7.5b), the two extra half-planes of atoms will align and become a complete plane.

These strain fields and associated forces are important in the strengthening mechanisms for metals.

04/19/23 15:57


12

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

04/19/23 15:57


13

Table 7.1: Slip Systems for FCC, BCC, and HCP Metals

04/19/23 15:57


14

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 normal to slip plane & stress direction

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

04/19/23 15:57


15

• 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

04/19/23 15:57


16Critical Resolved Shear Stress

R = 0

=90°

R = /2=45° =45°

R = 0

=90°

7.5 Slip in Single Crystals

04/19/23 15:57


17

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

And

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

Determining and angles for Slip in Crystals

For cubic xtals, angles between directions are given by:

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 R

equation to determine if slip is expected

1 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

u u v v w wCos

u v w u v w

04/19/23 15:57


18

EXAMPLE PROBLEM 7.1

Resolved Shear Stress and Stress-to-Initiate-Yielding ComputationsConsider a single crystal of BCC iron oriented such that a tensile stress is applied along a [010] direction.Compute the resolved shear stress along a (110) plane and in a direction when a tensile stress of 52 MPa is applied.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.

04/19/23 15:57


19

Stronger since 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 (slip) later.

04/19/23 15:57


20

7.6 Plastic Deformation ofPolycrystalline Materials

300 m

7.6 Plastic Deformation of Polycrystalline Materials

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:1. grain size reduction

2. solid-solution alloying

3. strain hardening

04/19/23 15:57


21

7.8 Strengthening by Grain Size Reduction

04/19/23 15:57


22

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:

04/19/23 15:57


23

21 /yoyield dk

7.8 Strengthening by Grain Size Reduction

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.

04/19/23 15:57


24

7.8 Strengthening by Grain Size ReductionGrain Size Reduction Techniques:

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

04/19/23 15:57


25


small impurities tend to concentrate at dislocations on the “Compressive stress side”

reduce mobility of dislocation increase strength

04/19/23 15:57


26

Large impurities concentrate at dislocations on “Tensile Stress” side – pinning dislocation


27


04/19/23 15:57

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

• Empirical relation:

• Alloying increases y and TS.

21 /y C~

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

04/19/23 15:57


28


7.10 Strain Hardening 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

04/19/23 15:57


29

Ti alloy after cold working:Dislocations entangle and multiplyThus, Dislocation motion becomes more difficult.

0.9 m

04/19/23 15:57


30

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 lengthunit volume

large hardening

small hardening

y0 y1

04/19/23 15:57


31


As cold work is increased Yield strength (y) increases.Tensile strength (TS) increases.Ductility (%EL or %AR) decreases.

04/19/23 15:57


32


• 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

Dd =12.2mm

Copper

04/19/23 15:57


33

What is the tensile strength & ductility after cold working to 35.6%?

% Cold Work

100

300

500

700

Cu

200 40 60

yield strength (MPa)

% Cold Work

tensile strength (MPa)

200Cu

0

400

600

800

20 40 60

340MPa

TS = 340MPa

ductility (%EL)

% Cold Work

20

40

60

20 40 6000

Cu

7%

%EL = 7%YS = 300 MPa

04/19/23 15:57


34

Cold Work Analysis

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

04/19/23 15:57


35

1 hour treatment at Tanneal...

decreases TS and increases %EL.

Effects of cold work are reversed!

3 Annealing stages to discuss...

Recovery, Recrystallization,and Grain Growth

ten

sile

str

eng

th (

MP

a)

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

04/19/23 15:57


36

Annihilation reduces dislocation density.

7.11 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

04/19/23 15:57


37

New grains are formed that:have a low dislocation densityare smallconsume cold-worked grains.

33% coldworkedbrass

New crystalsnucleate after3 sec. at 580C.

0.6 mm 0.6 mm

7.12 Recrystallization

04/19/23 15:57


38

• All cold-worked grains are consumed.

After 4seconds

After 8seconds

0.6 mm0.6 mm

04/19/23 15:57


39



04/19/23 15:57


40

TR

º

º

TR = recrystallization temperature

04/19/23 15:57


41

DESIGN EXAMPLE 7.1Description of Diameter Reduction ProcedureA 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.

04/19/23 15:57


42

At longer times, larger grains consume smaller ones.

Why? 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 no

n

coefficient dependent on material & Temp.

grain dia. At time t.

elapsed timeexponent typ. ~ 2

This is: Ostwald Ripening

04/19/23 15:57


43

Recrystallization Temperature, TR

TR = recrystallization temperature = point of highest rate of property change1. 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

04/19/23 15:57


44

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.

04/19/23 15:57


45

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

04/19/23 15:57


46

Coldwork Calc Solution: Cont.

For %CW = 43.8%

540420

y = 420 MPa– TS = 540 MPa > 380 MPa

6

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

04/19/23 15:57


47

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%

04/19/23 15:57


48

This process Needs an Intermediate 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:2 2

2 2

2 22 2

%% 1 100 1

100f f

s s

D D CWCW x

D D

0.5

2

2

%1

100f

s

D CW

D

22 0.5

%1

100

fs

DD

CW

0.5

1 2

200.30 1 0.335 in

100f sD D

Intermediate diameter =

04/19/23 15:57


49

Coldwork Calculations SolutionSummary: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

TSy

%CW1 10.335

0.4

2

x 100 30

Fig 7.19

04/19/23 15:57


50

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 size2. solid solution strengthening3. precipitate strengthening4. cold work

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

.

Summary

04/19/23 15:57


51

HomeWork Assignment 7, 12, 17, 24, 29, 41, D6 Due Monday 21/11/2011

04/19/23 15:57


52

Documents

Chapter 7: Dislocations & Strengthening Mechanisms 9/15/2015 11:21 PM Dr. Mohammad Abuhaiba, PE1