Principles of Materials Processing ──Technology of plas...

2016/4/6

1

Principles of Materials Processing ──Technology of plas city

LIU Juan

([email protected], 62933955, 13818928215)

Institute of Forming Technology & Equipment

SMSE SJTU

OutlineProcessing maps

Dislocation & plastic deformation

Plastic deformation of crystal

Elements of plastic theory

Recovery & recrystallization

Introduction

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Processing maps

Introduction

1

Plastic deformation of crystal

Elements of plastic theory

Recovery & recrystallization

1

1

1

General Review

Tensile Response of Materials

Effect of temperature & strain rate

1

Dislocation & plastic deformation 1

Reference books

Institute of Forming Technology & Equipment 4

• Mechanical Metallurgy *George E Dieter

McGraw‐Hill Book Company, London (1988)

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General Review


Mechanical Metallurgy

5

The central theme is structure ‐properties ‐ performance triangle.Changes in one are inseparably relatedto changes in the others; these changesare introduced by processing.

PROCESSING

STRUCTURE

PROPERTIESPERFORMANCE

mechanics

metallurgy

rationalizing

predicting

modifying

the response of metals to loads

PrincipleObjective

Problem

General Review


Overview of processes

6

Solidificationprocesses

Particulateprocesses

Deformation processes

Material removal

Heat treatment

Cleaning & surface treatments

Coating & deposition processes

Welding

Brazing & soldering

Adhesive bonding

Threaded fasteners

Permanent fastening methods

Manufacturingprocesses

Shaping processes

Property enhancing processes

Surface processing operations

Permanent joining processes

Mechanical fastening

Processing operations

Assembly operations

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General Review


Overview of metal forming

7

Metal forming: to change the shape of metal workpieces.

Plastic deformation: a permanent change of shape, i.e., the stress in materials

is larger than its yield strength.

Usually a die→ to force deformed metal into the shape of the die.

Metal forming is divided into:

(1) Bulk forming: large strain, small displacement, surface area to volume of

the work is small.

(2) Sheet forming: small strain, large displacement; Surface area to volume of

the work is large.

General Review


Bulk deformation processes

8

Rolling Forging

Extrusion Drawing

Traditionally Hot

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General Review


Sheet deformation processes

9

Bending Drawing

Shearing

Actually Cutting

General Review


Methodology

10

Two methods to study the strength, deformation and failure of materials :

(a) Engineering Mechanics or Continuum approach

‐ Assuming that materials are isotropic and homogeneous

‐ Applying mathematical methods, using global parameters to determine

stress state and material’s response to external forces.

(b) Materials Science or Microscopic approach

‐ Understanding of the mechanical properties based on the

atomic/mesoscopic scale.

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General Review


Methodology

11

Top‐down approach

Continuummechanics

Internal state variables

Constitutive equations

Bottom‐up approach

General Review


Important points to be kept in mind

12

Slip(Dislocation

motion)

Plastic Deformation in Crystalline Materials

Twinning Phase Transformation Creep Mechanisms

Grain boundary sliding

Vacancy diffusion

Dislocation climb

+ Other Mechanisms

Grain rotation

Phenomenological terms

No volume change

Shear stresses vs. Pure hydrostatic stresses

Fracture strain ← hydrosta c stresses.

slip vs. shear stresses at the slip plane.

Plastic deformation ↔ permanent deformation without external constraints

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General Review


Common modes of deformation

13

A body can be deformed using many modes:

Tension/Compression

Bending

Shear

Torsion Bending

Tension Compression

Deformed configuration

ShearTorsion

The Uniaxial Tension Test mechanical properties



The Uniaxial Tension Test (UTT)

14

Stroke →

Load

→

e →

s →

→

→

Gauge Length → L0 Possible axes

Tensile specimen

Initial cross sectional area → A0

Note that L0 is NOT the length of the specimen, but

the gauge length

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Stress and strain

15

Engineering Stress (s) and Engineering Strain (e) P↔ the load; ΔL↔ change in length.

Problems with the use of ‘s’ and ‘e’ (as outlined in the coming slides).

So, define True Stress () and True Strain ()

0A

Ps

0L

Le

0 → ini al

i→ instantaneous

Subscript

i

P

A

i

Le

L

Subscripts ‘0’ refer to initial values and ‘i’ to instantaneous values.



Problem with engineering stress and strain

16

Consider the following sequence of deformations:

L0

2L0

L0

e1→2 = 1

e2→3 = ½

e1→3 = 0

1

2

3

[e1→2 + e2→3] = ½

From stage 1 → 3 , no strain

Decomposition into 1 → 2 & 2 → 3 gives a net strain of ½.

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True stress and strain

17

iA

P

0

ln0

L

L

L

dLL

L

The definitions of true stress and true strain

Ai → instantaneous area

Same sequence of deformations considered before:L0

2L0

L0

1→2 = Ln(2)

2→3 = Ln(2)

1→3 = 0

1

2

3

[ 1→2 + 2→3] = 0

True strains are additive, engineering strains are not.



Schematic of s-e curve

18

O unloaded specimen

OY Elastic Linear Region (macroscopiclinear elastic region)

Y macroscopic yield point

The stress at this point is called yield strength

Note the increasing stress required for continued plastic deformation

(the stress to cause continued plastic deformation is called flow stress)

Neck

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19

YF Elastic + Plastic regime strain hardens flow stress increases with strain. be split into YN and NF

Neck

YN Stable region with uniform deformation along the gauge length

N Plastic Instability in tension Onset of necking




20

N True condition of uniaxialitybroken onset of triaxial state of stress (loading remains uniaxial but the state of stress in the cylindrical specimen is not).

Neck

NF most of the deformation is localized at the neck Specimen in a triaxial state of stress

F Fracture of specimen

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Information obtained from UTT

21

I. Young’s modulus*→ slope of the OY (elas c part of the curve).

II. Yield stress (or proof stress)→ stress corresponding to point Y.

III. Ul mate Tensile Stress (UTS) → point N (maximum) in s‐e plot.

IV. Fracture stress→ stress corresponding to point F.



Information obtained from UTT

22

IV. Ductility→ measured as: (a) strain at fracture (in %), (b) % reduction in area.

V. Resilience (area under the curve elastic portion‐ OY).

VI. Toughness (area under the curve total)→ has unit of Energy/volume [J/m3].

VII. Strain hardening exponent (from ‐ plot).

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Q & A

23

What is meant by toughness?

The energy absorbed by the material up to failure.

By a uniaxial tension test or in an impact test (using a notched specimen).

Combined parameter involving strength and ductility.

What are the simple tests to evaluate their mechanical behaviour?

Uniaxial tension test.

Compression test.

Hardness test.

Bending test (3‐point, 4‐point bend tests).

Torsion test.



Q & A

24

Why does the stress value drop after UTS in a “s‐e” plot?Why does this not happen in the “‐” plot?

After necking the load bearing area decreases,

Computation of ‘s’ using A0 → leading to a apparent drop in stress

‘apparent’ in reality the material continues to harden

In the “‐” plot using the instantaneous area

the hardening effect with deformation captured (even beyond the necking)

the necking event cannot be ‘captured’

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Q & A

25

What is the difference between Stiffness and Young’s modulus?

Both relate to the elastic property of the material.

Stiffness: the structure/geometry (of sample) dependent property

e.g. the stiffness of a spring

Young’s modulus : a geometry independent material property

e.g. Young’s modulus of polycrystalline Cu



Result of an actual stress‐strain plots

26

Specimen ID Aluminium

Cross Sectional Area 30.2 mm2

Gauge Length 28.44 mm

Cross head velocity 2 mm/min

Test mode Stroke

Peak Stress 193.0 MPa

Peak Load 5.83 kN

Specimen Mild Steel

Area 27.91 mm2

Gauge Length 28.24 mm

Cross head velocity 2 mm/min

Test mode Stroke

Note the ‘serrated’ yielding (yield point phenomenon)

Peak Stress 595.7 MPa

Peak Load 16.6 kN

• not good to calculate Young’s modulus from s‐e plot

• better to determine the Young’s modulus from sound propagation experiments.

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Comparison between “Engineering” and “True” quantities

27

Comparison between true strain and engineering strain

True strain () 0.01 0.10 0.20 0.50 1.0 2.0 3.0 4.0

Engineering strain (e) 0.01 0.105 0.22 0.65 1.72 6.39 19.09 53.6

Note that for strains of about 0.4, ‘true’ and ‘engineering’ strains can be assumed to be equal. At large strains the deviations between the values are large.

iA

P

0

ln0

L

L

L

dLL

L

0

ln 1 1 ln(1+e)L

L

0

0 0 0

1 1 (1 )i i

i

A L LPs s s e

A A L L

00 0 i i

0

From volume constancy A L =A L i

i

A L

A L

Valid till necking starts



Yielding

28

Microscopically, dislocations leave the crystal (grain) and cause microscopic plastic deformation (of unit ‘b’) determined from microstrain (~10–6 ) experiments on single crystals.

True elastic limit (microscopically and macroscopically elastic→ not microscopic yielding) ~10–6 [OA portion of the curve]

Microscopically plastic, macroscopically elastic→[AY portion of the curve]

Where does Yielding start?

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Yielding

29

Offset Yield Strength (proof stress) parallel to the elastic line at a given strain like 0.2% (= 0.002)

In some materials (e.g. pure annealed Cu, grey cast iron etc.) at some given strain (0.005).

What is offset Yield Strength (proof stress)?

The stress‐strain behavior or brass



Key points for yielding stress

30

y : yield stress in an uniaxial tension test.y should not be used in other states of stress.

I.e in uniaxial tension the yield criterion is very simple:

Yielding starts when: y .

Hydrostatic pressure does not lead to yielding in a continuous solid (usually!).

This implies that the stress deviator holds the key to yielding.

For an isotropic material the yield criterion will be a function of the invariants of

the stress deviator.

Two commonly yield criteria are:

Von Mises or Distortion‐Energy Criterion

Maximum shear stress or Tresca Criterion.

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Ductility

31

Slip is competing with other processes leading to failure.

A ductile material yields before failure (i.e. y < f).

Ductility depends on the state of stress.

How to measure the ‘ductility’ of a material from the uniaxial tension test? Strain at fracture (ef), (often called elongation, a dimensionless quantity) Reduction in area at fracture (q)

‘q’ is a better measure of ductility as it does not depend on the gauge length (L0);.

0

0

(%) 100ff

L Le

L

0

0

(%) 100fA Aq

A



Comparison between reduction in area versus strain at fracture

32

Two measures of ductility: Strain at fracture (ef) Reduction in area at fracture (q)

Representing different aspects of material behaviour.

‘ef’ : controlled by uniform elongation prior to necking ;

dependent on the strain hardening capacity(more the strain hardening, more will be the ‘e’).

‘q’ (area based calculation) comes from the necking process (which is more geometry dependent).

0

0

ff

L Le

L

0

0

fA Aq

A

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Necking

33

What happens after necking?

Till necking the deformation is ~uniform.

Till necking points on the ‐ plot lie to the left and higher than the s‐e plot. After necking deformation is localized around the neck region.

i.e. Necking : an instability in tension.

Fractured surfaces

Neck

What happens after necking?



Ductile material

34

A ductile material : y < f (i.e. on loading it yields before fracture).

(Microscopically, in yielding we drive dislocations,

in fracture we propagate cracks).

Two factors contribute to this fracture.

(i) Necking:

Which reduces the load bearing area

Which introduces a triaxial state of stress

(triaxial state of stress tends to suppress plasticity by slip).

(ii) work hardening, which makes y > f

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Ductile vs. Brittle Behavior

35

Brittle

Moderate Ductility High

Ductility

The stress‐strain behavior of brittle materials compared with that of more ductile materials.



Classifications of Stress-Strain Plots

36

The types of curves reflect different deformation characteristics

Type I completely elastic behavior Type II elastic behavior and homogeneous plastic deformation.

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Classifications of Stress-Strain Plots

37

Type III elastic behavior followed by heterogeneous plastic flow caused by twin controlled deformation or solute atom‐dislocation interactions.

Type IV exhibiting a narrow heterogeneous deformation region between initial elastic and final homogenous flow regions.



Hollomon’s Equation

38

Consider the true stress‐strain curve of a metal beyond yielding, where most ofthe strain is plastic strain.

Where, ‘n’ is the strain hardening exponent and ‘K’ is the strength coefficient. At low temperatures (below recrystallization temperature‐ less than about 0.5Tm)

strain hardening is very important to obtain good ductility. During tensile deformation instability in the form of necking localizes

deformation to a small region (a triaxial state of stress). In the presence of strainhardening the neck portion (which has been strained more) hardens and thedeformation is spread to other regions, thus increasing the ductility obtained.

nK

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Hollomon’s Equation

39

,

n

TK

When true strain is less than 1, the smaller value of ‘n’ dominates over a larger value of ‘n’

K → strength coefficientn → strain/work hardening exponent◘ Cu and brass (n ~ 0.5) can be given large plastic strain (before

fracture) as compared to steels with n ~ 0.15.

Material n K (MPa)

Annealed Cu 0.54 320

Annealed Brass (70/30) 0.49 900

Annealed 0.5% C steel 0.26 530

0.6% carbon steel Quenched and Tempered (540C)

0.10 1570

‘n’ and ‘K’ for selected materials,

ln

ln T

n



Average flow stress

40

• For many bulk deforming processes, rather than instantaneous stress, average stress is used (extrusion)

• The average flow stress can be obtained by integrating the flow stress along the trajectory of straining, from zero to the final strain value defining the range of interest

n

kY

n

f

1

Average flow stress

Max. strain during deformation

Strength Coefficient

Strain hardening exponent

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Example

41

Determine the value of the strain‐hardening exponent for a metal that will cause the average flow stress to be three‐quarters of the final flow stress after deformation.

According to the statement of the problem, we have

4/3fY of fY

333.0

75.0)1/(1

75.0)1/(

75.0

n

n

KnK

YYnn

ff

Effect of temperature & strain rate on flow properties


Flow characteristics

42

,

m

TC

C → a constantm → index of strain rate sensitivity

◘ If m = 0 stress is independent of strain rate (stress-strain curve would be same for all strain rates)

◘ m ~ 0.2 for common metals◘ If m (0.4, 0.9) the material may exhibit superplastic behaviour◘ m = 1 → material behaves like a viscous liquid (Newtonian flow)

The effect of strain rate is compared by performing tests to a constant strain

At high temperatures (above recrystallization temperature) where strain rate is the important parameter instead of strain, a power law equation can be written.

,

ln

ln T

m

Thermal softening coefficient ()

ln

lnT

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43

Slip is a thermally activated process. So increasing the temperature decreases the

flow stress.

Increasing the strain rate is like decreasing the temperature. The strength of the

material increases on increasing the strain rate (with decrease in ductility)

The effect of these two parameters on the stress‐strain curve is slightly different

for FCC and BCC metals.




44

What is the importance of ‘m’ and ‘n’ (strain hardening exponent and strain rate sensitivity)?

Below recrystallization temperature ‘n’ is ‘the’ important parameter.

Above recrystallization temperature it is ‘m’ which is important.

It is necking which limits the ductility in uniaxial tension.

Necking implies that there is locally more deformation (strain) and the strain rate is also higher locally.

Hence, if the ‘locally deformed’ material becomes harder (stronger) then the deformation will ‘spread’ to other regions along the gauge length and we will obtain more ductility.

Hence having a higher value of ‘n’ or ‘m’ is beneficial for obtaining good ductility.

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Flow stress modelling

45

At the high stress:

Sellars :

exp QZ fRT

2

3 sinhn

f A

11

nf A

2 expf A

At the low stress :

The peak stress modelled using creepequation

Clarence�Zener(1905-1993)

s

ss

c p

0

c p

Z: the temperature‐compensated strain rate,i.e. the Zener‐Hollomon parameter; isstrain rate; Q is the effective activationenergy for deformation; R is the gasconstant; T is temperature; σ is the stress



Flow characteristics for AZ61

46

10.001s 10.01s

10.1s 11s

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00

20

40

60

80

100

120

140

160

180

200

673 K623 K

573 K

Tru

e st

ress

,

(MP

a)

True strain,

523 K

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00

20

40

60

80

100

120

140

160

180

200

673 K623 K

573 K

Tru

e st

ress

, (

MP

a)

True strain,

523 K

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00

20

40

60

80

100

120

140

160

180

200

673 K623 K

573 K

Tru

e st

ress

, (

MP

a)

True strain,

523 K

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00

20

40

60

80

100

120

140

160

180

200

673 K623 K

573 K

Tru

e st

ress

, (

MP

a)

True strain,

523 K

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47

3 5 7

sinh2 3! 5! 7!

x xe e x x xx x

0.5x sinh x x

2x sinh 2xx e

0 0.5 1 1.5 2 2.50

1

2

3

4

5

6

7

xf(

x)

f(x)=sinh(x)f(x)=x

f(x)=ex/2

(sinh(x))n xn (exp(x)/2)n E1 E2

x=0.5 0.5211 0.5 0.8244 4.05% 58.20%

x=2.0 3.6269 2 3.6945 44.86% 1.87%

1

sinh100%

sinh

n n

n

x xE

x

2

sinh100%

sinh

n nx

n

x eE

x



Peak stress modelling

48

1nf A

nZ A

ln ln lnZ A n

exp nQ ART

ln ln lnQ n ART ln

ln T

n

ln1

Q nR

T

At constant temperature:

At constant strain rate:

ln ln1 ln T

Q R

T

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49

2 expf B

expZ B

ln lnZ B

exp expQ BRT

ln lnQ BRT ln

T

1Q

R

T

ln1 T

Q R

T

At constant temperature:

At constant strain rate:




50

3 sinhn

Z C 3exp sinhnQ CRT

At the high stress:

11

nf A

2 expf A

At the low stress :

ln ln Plotting 1n

Plotting

ln 3n

1ln sinh

T

Plotting

Plotting ln ln sinh

ln sinh ln1 ln sinh

T

Q R

T

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Example No. 1

51

For Magnesium alloy AZ61:

MPa 250℃

(513K)

300℃

(553K)

350℃

(593K)

400℃

(633K)

0.001 110 75 52 38

0.01 130 95 62 50

0.1 150 122 95 75

1 162 138 115 97

Peak stress for different temperatures and strain rates



Example No. 1

52

-8 -7 -6 -5 -4 -3 -2 -1 0 13.0

3.5

4.0

4.5

5.0

5.5

523K 573K 623K 673K

ln p

ln. -8 -7 -6 -5 -4 -3 -2 -1 0 1

0

40

80

120

160

200

523K 573K 623K 673K

p

ln.

Temperature, (K) n1 Mean value of n1

523 17.6518

11.0209573 11.0731

623 8.2007

673 7.1579

Table The values of n1 at different temperatures.

Temperature, (K) β Mean value of β

523 0.1308

0.1138573 0.1066

623 0.1037

673 0.1140

Table The values of β at different temperatures.

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Example No. 1

53

2exp exp pQ ART

2ln ln pQ ART

ln1

p

p T

Q R

T

0.0014 0.0016 0.0018 0.00200

40

80

120

160

200

0.001/s 0.01/s 0.1/s 1/s

p

1/T.

Q equals 163.26KJ/mol.

Strain rate, (1/s) əσp/ə(1/T) Mean value of əσp/ə(1/T)

1 1.7001e+005

1.7256e+005

0.1 1.8916 +005

0.01 1.7757e+005

0.001 1.5351e+005



Example No. 1

54

20 22 24 26 28 30 32 34 36 38

40

60

80

100

120

140

160

180

200

lnZ

Experimental data Linear fit

p

2

1lnp

Z

A

β is 0.1114, A2 is 1.4161×108.

0 10 20 30 40 50 60-4.0

-3.5

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

lnZ

Experimental data Linear fit

ln p

1 2ln lnp Z B B

B1 is 55.7601, B2 is 12.9082

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Example No. 2

55

For Magnesium alloy AZ31:

MPa 250℃

(513K)

300℃

(553K)

350℃

(593K)

400℃

(633K)

0.001 100 64 45 30

0.01 122 85 60 42

0.1 150 112 84 60

1 182 150 110 82

Peak stress for different temperatures and strain rates



Example No. 2

56

n

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Example No. 2

57

The approximate values of αand Q equal 0.0113 and158.7323KJ/mol.

ln sinh ln

1 ln sinhT

Q RT



Example No. 2

58

11 2 2

1ln 1

n n

p

Z Z

A A

1 2ln lnp Z B B

Where: n=6.178, A=9.8194E11, B1=47.5013, B2=9.7151

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59

ss

0

cp

s

c p

WH s ss

1 2

d

dk k

Work hardening and DRV:

DRX:

2

20

k

WH s s e

1 expdn

cdrx d

p

X k

1 expdn

cWH s ss d

p

k

2

20

1 expd

k

WH s s c

n

cWH s ss d c

p

e

k




60

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

20

40

60

80

100

120

140

160

0.001s-10.01s-1

0.1s-1

True strain

Tru

e st

ress

(M

Pa)

Experiment data Calculated result

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

20

40

60

80

100

120

140

160

0.001s-10.01s-10.1s-1

True strain

Tru

e st

ress

(M

Pa)


0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

20

40

60

80

100

120

140

160

0.001s-10.01s-10.1s-1

True strain

Tru

e st

ress

(M

Pa)


0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

20

40

60

80

100

120

140

160

0.001s-10.01s-10.1s-1

True strain

Tru

e st

ress

(M

Pa)


Comparsion of the predicted results of the 30Cr2Ni4MoV steel (a) T=970oC (b) T=1070 oC (c) T=1170 oC(d) T=1270 oC

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Thank you !

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