1

How to Make Strong Metals

With High Ductility?

Yan Beygelzimer

Donetsk Institute of Physics and Technology National Academy of Sciences of Ukraine

2

Outline

1. How do we try to do it?

First way: fragmentation

Second way: consolidation

2. How do we try to explain and to predict it?

Mechanics of the processing

Structure, ductility, strength

3. Final Thoughts

3

How do we try to do it?

•Severe plastic deformation (SPD)

•Twist Extrusion

•First way: Fragmentation

•Second way: Consolidation

4

High Pressure Torsion Equal Channel Angular Extrusion

Some of the SPD Techniques

3d Forging Twist Extrusion(Y.Beygelzimer, 1999)

5

Twist channel

The idea of TE:

6

Twist channel

Equivalent strain e1

The shape and the dimensions of the work-piece do not change!

The idea of TE:

7

Twist channel

Equivalent strain e2

The idea of TE:

Repeated twist extrusion leads to grain refinement

8

Twist Extrusion: Two in One

Twist Extrusion

Consolidation

Fragmentation

9

•Grains refinement. UFG materials

Fragmentation

•Breaking of a brittle particles

10

Why should we care about grain refinement during severe plastic

deformation?

This is one of few effective techniques for obtaining ultrafine-grained (UFG) materials

11

SPD

Grain refinement

Coarse

12

SPD

Grain refinement

UFG

13

Ultrafine-Grained Materials

What are they?– Metals with grain size ~10-1000 nm

Why are they appealing?– Significantly improved properties– Qualitatively different properties not seen

in conventional materials

14

High strength and ductility of UFG materials

R.Z. Valiev, I.V. Alexandrov, Y.T. Zhu and T.C. Lowe, “Paradox of Strength and Ductility in Metals Processed by Severe Plastic Deformation,” J. Mater. Res. 17, 5-8 (2002).

!

!

15

Twist extrusion of the Ti

50 m 50 m

Initial state After 3 passes

16

Twist extrusion of the Ti

17

SPD of the Ti

18

Plates for traumatology are made of UFG Ti

19

Twist extrusion of the Al-Mg-Sc-Zr alloy

Chemical composition: Al - 3 wt.%Mg - 0,3 wt.%Sc – 0,10 wt.%Zr

Twist extrusion: T=280-300ºC 5 passes CW + 1 pass CCW

dav=0.325 µm dmin= 0.077µmdmax=0.671 m

Initial grain size dav=100 µm

Standard direct extrusion:T=280-300ºC

dav= 0.455 µm dmin= 0.129µmdmax=1.032 m

20

SPD of the Cu

440

15440420ECAE, 16 passes,

without BP

18426393TE, 3 passes,with BP

Uniform elongation,

%Tensile strength

MPaYield stress,

MPa State

4014580As received

14360320ECAE, 2 passes

25470ECAE, 16 passes,with BP

*

*

*

*N. Krasil’nikov, R.Valiev

21

ECAE of the Cu

Wei Wei, Guang Chen, Jing Tao Wang, Guo Liang Chen

Journal of Advanced Materials 2005 (in press)

22

Three present-day ways to increase strength and ductility of UFG materials

•Cryogenic processing to produce a bimodal grain structure

•Formation of second-phase particles to modify the propagation of shear bands

•Developing aging and annealing treatments that can be applied to the UFG materials in the post-processed condition

23

SPD

Breaking of a brittle particles

24

SPD

Breaking of a brittle particles

25

Al-88%; Si-9,5% …

σy=50МPа

Initial state 1 pass

δ=1% σy=205МPа δ=14%

TE of secondary Al alloy

26

Twist Extrusion of phosphorous Cu (P 9%)Initial state 1 pass

YS,MPa

11420Twist Extrusion

4360Direct Extrusion, µ=4.5

, %YS, МPаTreatment

Back pressure = 200 MPa, T= 623 K

27

SEM, As-deformed structure,TE – 5 passes, Tdef=280-300ºCLongitudinal section

Twist extrusion of the Al-Mg-Sc-ZR alloys

Chemical composition: Al – 3 wt.%Mg - 0,3 wt.%Sc – 0,15 wt.%Zr

SEM, As-cast structureCross-section

0.5µm0.5µm

28

Consolidation

SPD

29

Consolidation

SPD

30

Consolidation of nanostructural Cu powder by Twist Extrusion

2 passes1 pass

Compression test after 2 passes:

Yield stress = 450 MPaBreaking strain = 28%

Tensile test after 2 passes:

Yield stress = 200 MPaBreaking strain = 15%

Initial powder, D=250 µm

Back pressure = 200 MPa, T= 473 K

31

Consolidation of nanostructural Cu powder by Twist Extrusion

StateDensity,

%

Diameter of the coherent-

scattering region L, nm

powder - 100

TE, 1 path 99.2 36

TE, 2 paths 99.6 55

32

Microhardness and the volume fraction of the amorphous phase in the compacted samples.

X-ray diffraction patterns

Consolidation of amorphous Al86Ni6Co2Gd6 melt-spun ribbons by Twist Extrusion

300 350 400 450 500 550350

400

450

500

550

0,0

0,2

0,4

0,6

0,8

1,0

Mic

roh

ard

nes

s (k

gf/

mm

2 )

Extrusion temperature (K)

Al86

(Ni,Co)8Gd

6

Ribbons

Extrusions Vo

lum

e d

ensi

ty o

f cr

ysta

ls

20 40 60 80 100 120 140

Inte

nsity

(a.u

.)2 (deg)

AQ513 K

573K

33

Yield stress =180-220 МPа; = 20-24%

d=12 mm

Consolidation of the cutting of the secondary Al alloy by Twist Extrusion

34

How do we try to explain and to predict it?

•The questions

•Mechanics of the processing

•The Problem

•The Model

•Predictions

35

•Parameters of the process?

•Strength and Ductility of the materials?

•Structure of the materials?

The questions

36

Mechanics of metal flow

37

Stream lines

4

5

6 7 8

1

2

3

38

Kinematically-admissible velocity field

0Vdiv

0nV

0VV

1. Volume constancy condition

2. Boundary condition

39

Kinematically-admissible velocity field

ArotV

);,,(),,(

;5.0

;5.0

0

0

0

zyxPzyxxyR

tgVA

xVA

yVA

z

y

x

0

0

0

P- function which is varied

- form function

40

Finding function P on the experimental stream lines

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36

0

3

3

z( )

p z( )

360 z

Experimental stream line

Theoretical stream line

41

Metal Deformation under Twist Extrusion

We showed that most of the deformation achieved by Twist Extrusion is Simple Shear at the ends of the twist channel

42

Equivalent strain for TE pass

The average equivalent strain during one pass e=tan()

-15-10

-50

510

15 -8-6

-4-2

02

46

80

100

200

300

400

500

Z A

xis

Y AxisX Axis

Yield stress, Cu

Distribution of the strain

43

The Problem with Theoretical Model

One of the main problems faced by any theoretical model is the need to capture the multi-level character of plastic deformation

44

Metal Structure is Determined by the Image of the Loading Process

e2

e1

P1,T1,...

P2,T2,...

P3,T3,...

P4,T4,...

Gra

in r

efin

emen

t

45

The reason is that the structure defines the mechanical

properties of the materials.

But … The Image of The Loading Process depends on this structure

46

Macro-Micro Interdependency

Image of the Loading Process

Metal Structure

47

The Problem

• We are trying to produce a given ultrafine homogeneous structure

• In reality, however, the specimen may respond with a number of bad things: highly inhomogeneous structure, deformation localization or fracture.

• The reason why this happens is precisely the “Interdependency”.

48

Continuum model

FEM

Our approach to capture the interdependency

Internal parameters

Internal parameters will allow us to account for the interdependency between the stress-strain state and the structure.

49

S

Internal parameters serve as special envoys representing the micro-level

processes at the macro-level

50

Model of the material

51

Constitutive equations of theMises’s model

ikik3

1

ij

ijijij

fef

,0

ikikikik

3

1

3

1

2

2

3

sijf

0e

RVE

52

Porous body with structurally inhomogeneous matrix

2

2

3

sijf

0e

RVE

Beygelzimer Y. et al., (1994) Engin. Fracture Mech., 48, N5

Beygelzimer Y. Proceedings of International Workshop on Modelling of Metal Powder Forming Processes, Grenoble, France, July 21-23, 1997

222

31

s

ijf

01 ke

53

Plausible reasoning

2

2

3

sijf

RVE

222

31

s

ijf

54

Loading surfaces

= 0

55

Loading surfaces of the cutting of the secondary Al alloy

0 100 200 300 400 500 600 700 800

50

100

150

P,M Пa

, M Пa

1 2

3

4

: 1- 30%, 2- 20%, 3- 10%, 4- 3%.

, MPa

P, MPa

56

Porous body with structurally inhomogeneous matrix at <<1

2

22

36

sij af

s

ad

d

36

a,s (d)

57

Beygelzimer Y., Shevelev A. On the Development of Fracture Models for

Metal Forming// Russian Metallurgy (Metally), Vol., N5, p. 452–456, 2003

Breaking of a brittle particles

)exp(1)exp( 21

58

Beygelzimer Y. Grain refinement versus voids

accumulation during severe plastic deformations of

polycrystals: Mathematical simulation, Mechanics of

Materials, V. 37, N7, p. 753-767, 2005

The model of grain refinement and viscous fracture

59

Main characters of the Model

0

500 nm

• Accumulative Zone – a “spring”, the part of crystals in which dislocation charges accumulate during plastic deformation; AZs emerge due to the inhomogeneity of shear along the sliding plane.

• Void – a bit of an emptiness

• Embryo – an embryo of high-angle boundary (a partial disclination)S

cale

60

Pictures of Main Characters

61

The Birth of an Accumulative Zone

There are regions of polycrystals where dislocations get plugged during plastic deformation. Such regions cause bendings of the crystalline lattice.

62

The Birth of an Accumulative Zone

The model postulates that AZs emerge in two places:

1. hurdles that exist in polycrystals before deformation

2. high-angle boundaries that emerge during deformation.

63

Relaxations of Accumulative Zones in

coarse grained materials

There are different relaxation mechanisms for accumulative zones. When talking about large cold deformations, we will distinguish two main mechanisms:

1. Emergence of high angle boundaries (leading to grain refinement)

2. Emergence of voids (leading to fracture)

64

Relaxations of Accumulative Zones in

coarse grained materials

65

Boundary sliding

0 0.1 0.2 0.3 0.4 0.51

0.5

0

0.5

1

1.5

f1 s( )

f2 s( )

f3 s( )

s

1b

2b

f

1bs

99.02

1 b

b12

1 b

b80.0

2

1 b

b

f

boundary1b

2b

Prandtl model

66

d

Sliding plane

Relaxations of Accumulative Zones in

ultrafine grained materials

67

Relaxations of Accumulative Zones in

ultrafine grained materials

68

Relaxations of Accumulative Zones in

ultrafine grained materials

69

Equations of the Model

S

Constitutive

G(,)=0 P(,;µ)=0

General

Classical Plasticity Theory:

Proposed model with internal parameters capturing the structure:

P(,; S,; µ)=0

G(,)=0

K(S,, ,;µ)=0

General

Constitutive

Kinetic

70

Loading function

vdNC c32

3

3*

2

*22

36,;

SaSf s

ij

71

Kinetic equations

6

32

3

3

0

5

54

43521

CvdNCd

d

SS

NC

d

Sd

NCNCd

Nd

NCCSFNCCCd

Nd

c

b

bb

b

0 .

0N 0bN

0S 0

0N 0bN

0S 0

01 nN nbnb NN 1

nn SS 1 nn 1

Quasimonotonic loading

Cyclic loading

72

Prediction

73

As :

,0N ,0bN

,cdd 0

Metals don’t fracture and don’t harden under sufficiently

high pressure when equivalent strain is very large, i.e.,

metals become ideally plastic.

Ideal Plasticity Land

Prediction (i)

The reason is boundary sliding

74

Ductility grows for sufficiently large equivalent strain

Prediction (ii)

The reason is boundary sliding

75

p

Bd

d * vdNC c32

3

3*

0 5 100

0.005

0.01

0.015

strain

alph

a

*с

The term responsible for the birth of voids

Ductility grows for sufficiently large strain. α* decrease for sufficiently large strain.

76

P

Grain refinement intensity grows and fracture decreases with the increase of pressure in the center of deformation.

Prediction (iii)

77

p, MPa0.1 420 840 1120

S

3

1

0

с

pс

D

pN

The reason is p grows with pressure increase

78

Correspondence with experimental results: Influence of pressure on the microstructure of

molybdenum at hydroextrusion:

Pb = 0,1 MPa Pb = 800 MPa

e=0.36

e=0.60

e=0.92

30m

Pb

79

Effect of pressure on fragment size distribution Hydroextrusion of molybdenum (e=0.6).

40

20

f,%

l, 100 µm

, GPa

Correspondence with experimental results:

80

Twist Extrusion based on mechanical extrusion with backpressure

Twist Extrusion based on hydro-mechanical extrusion

Use in Twist Extrusion

81

- Рb 0.1 MPa

- Рb = 1000 MPaP

P

Model-based prediction for Twist Extrusion

82

Grain refinement intensity is typically higher for

simple shear than for uniaxial elongation

Prediction (iv)

To get intense grain refinement, one needs to choose

deformation schemes with small value of ductility

and to perform deformation under pressure.

83

p3

Ductility diagram, steel 08X18H10T (Experiment data V.Kolmogorov, et al., 1986)

c

c(0), torsion

c(0), tension under pressure

Sс(0)

S

3

1

0

с

pс

D

pN

The reason is simple shear has

smaller ductility (under the same

pressure).

c(0) is the metal ductility at p=0

84

Ductility diagrams for various metals show that, as a rule, ductility с(0) of tension greater than that of torsion.

(Experiment data V.Ogorodnikov and I.Sivak, 1999)

85

Quasi-monotone deformations provide higher grain refinement intensity than cyclic deformations.

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

strain

N, 1

/mkm

^2

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

strain

S, 1

/mkm

N, m

-2

S, m

-1

Prediction (v)

86

Cyclic deformations provide higher ductility than quasi-monotone deformations

0 0.1 0.2 0.3 0.4 0.50

0.001

0.002

0.003

strain

poro

sity

C5

por

osi

ty

87

In order to increase the intensity of grain refinement under cyclic deformation, one has to increase the amplitude of deformation.

For example, in Twist Extrusion we combine clockwise and counter-clockwise dies.

88

clockwise

clockwise

A

89

clockwise

counter-clockwise

2A

90

To avoid cyclic deformation we combine Twist Extrusion with Spread Extrusion

Spread ExtrusionTwist Extrusion

91

P P

Layers of the voids

Prediction (vi)Sufficiently high pressure in the center of deformation prevents strain localization

92

Prediction (vii) Grain size distribution

0 1 2 30

1

2

Sdf(

d)/S

Self-similar stage

0 5 10 150

0.5

d, mkm

f(d)

,1/m

km

3

2

1

md ,

1,m

f

1,m

f

md ,

0,5

When there are fragments with size dc

dc

93

Self-similarity of Experimentally Obtained Distributions

.

0 0.5 1 1.5 2 2.50

2

4

6

d, mkm

f, m

km^(

-1)

Cu, torsion(V.Panin et al., 1985)

93%

46%

0.28

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

d/M(d)

fxM

(d)

md ,d

d

df =4.61

1.07

1,m

f

94

.

During the self-similar stage of grain refinement, the fragment boundary mesh in the cross-section of the specimen represents a fractal set with dimension , 1< <2.

S 11d

1< <2 - fractal dimension.During the self-similar stage is constant

Prediction (iv)

95

.

s S

s d

1

were ,1 10 During the self-similar stage is constant

Hall-Petch

96

.

When sufficiently many indivisible fragments of size dc appear, the self-similarity of the boundary mesh gets violated. In this case

Sd

1

sd

1

97

Simulation of the grain refinement processes by Cellular Model

Initial grain

Plastic deformation

Deformation continues. Rotation moments appear

М М

Moment stresses relax through grain splitting

RVE initial state

RVE new state

Outer stress 0

Stress on each element n

Extension Dnt and rotation nt of each element

Beygelzimer Y., et al., Philosophical Magazine A, 79, N10, (1999)

98

Grain size, μm

Limiting Grain size distribution (Cellular Model)

10µm

Cu

99

Final Thoughts

100

S

Continuum model

FEM

One can substitute the classical plasticity model by the above model in any FEM package to directly compute the stress-strained state of a metal and its interdependence with the structure

What do we hope for?

101

What do we hope for?

102

• The model is relatively new

• The limits are not entirely explored

• A long way toward good parameter estimation

… but there are grounds for hope

What do we have?