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Natural Strain Near Welding Interface For Different Collision Angles

EPNM-2008 Lisse, Netherlands

H.H. Yan; X.J. Li

May 6-9, 2008

Dalian University of Technology

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Outline of talk

• Introduction

• Model

• Solutions

• Results

• Discussions

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Introduction

• Where will strain rate and strain be used?• To analyze the acting of mechanics and

thermodynamics• To setup the model of heat transmission• To estimate the thickness of layer melted• And so on

Temperature ?For example:

Base plate

Flyer plate

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Temperature?

Caused by Shock compaction

Caused by deformation energy

dtYdYdUU peq

peq 00

2 2 222 3 2

6 ( )3 2 23

xypx xxyeq

Strain rate?Strain?

VC

UT

0

To research here

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Model

Ideal fluid

Symmetrical collision

1

6

Reason of adopting the model

Detonation of explosive is about 2000~3000 m/s.

Pressure (1.6~3.5) × 1010 Pa

bfv 2/2b

Strength of materials >10

bnegligible

So

compressibility small

+

<6%

IdealLiquidmodel

7

Fig. Plane diagram of conjugate

complex velocity

Ψ1

0

Vf? B

u

B’

C’

D’

D

A

A’

C

v

Ψ2

Ψ3

Ψ4

Transforming

holographic function method

2

fV

H

21

1 ；fV

H

21

2 ；fV22H

3 ；fV

H

22

4

iyxz yxiyxzw ,,

Boundary condition:

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Solutions (Strain)

• Strain rate

• Strain

X Horizontal ordinateY Vertical ordinateu Horizontal velocityv Vertical velocity

x

u

,y

u

,x

v

,y

v

x

u

y

uy

u

x

u

E

Eddy factor considered tnn

Tn EEE

1

Along stream line

Points on the line

Steady assumptionIncompressiblityconservation of momentum

model

holographic function method

finite difference method

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solving x,y,u,v• The complex potential yxiyxzw ,,

yx,

yx,

velocity potential

stream function

1 1 1 1

2

2 2 2 2

2

2 2 2 2 1 1

1( , ) 2[(1 cos ) (1 cos ) ] /( (1 cos )) (3.1)

/ 2 1 1

{(1 cos ) ln[(1 ) ] (1 cos ) ln[(1 ) ]/ 2

cos {ln( ) ln( )} 2 sin { }} /( (1 cos )) (3.2 )

f

v v N Ku v tg tg tg tg

V H u u M J

xu v u v

H

N KM N J K tg tg a

M Jy

H

1 1 2

2

2 2 2 1 1

12{ (1 cos ) (1 cos ) sin {ln[( cos )

/ 2 1 1 2

( sin ) ] ln[( cos ) ( sin ) ]} cos { }} /( (1 cos )) (3.2 )

v vtg tg u

u u

N Kv u v tg tg b

M J

Xyuv

solved

For stream line given

With u,v changing

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Stream lines plotted

Fig. Different flow diagrams with differentψ at

β = 130

Y(H

2/2

)

X( H2 /2)

~ ＝ 0.5

~ ＝0

~ ＝- 0.5

~ ＝-1

~ ＝-2

3

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Solving stain rate (along stream line)

x

u

y

uy

u

x

u

E

0

y

v

x

u

incompressibe

0

y

u

x

v

irrotational

yx y

v

x

u

y

u

x

v

y

uxy

2

)/()(2

)/(2

43124321

4312121

ffffffHH

fffffHH

xy

yx

)sincossincos

(cos})sin()cos(

sin

)sin()cos(

sin{sin

)1(

1)cos1(

)1(

1)cos1(

222222

2222221

KJ

KJ

NM

NM

vu

v

vu

v

vu

u

vu

uf

)sincoscossin

(sin})sin()cos(

1

)sin()cos(

1{cos)cos(

)1(

1)cos1(

)1(

1)cos1(

222222

2222222

KJ

JK

NM

NM

vu

vuu

vu

u

vu

uf

)sincossincos

(sin})sin()cos(

sin

)sin()cos(

sin{cos

)1()cos1(

)1()cos1(

222222

2222223

KJ

KJ

NM

NM

vu

v

vu

v

vu

v

vu

vf

)sincoscossin

(cos})sin()cos(

1

)sin()cos(

1{sin)cos(

)1()cos1(

)1()cos1(

222222

2222224

KJ

JK

NM

NM

vu

vuu

vu

v

vu

vf

solved

second-order tensor

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Strain rate results (For example)

-10 -8 -6 -4 -2 0 2 4 6 8 10-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

Fig. Diagram of strain rate with different angles at ~ ＝ -0.2

x

xy

Ⅱ

Ⅰ

0/ xstrainrate。

X/(H2/2

)

-1.60 -1.55 -1.50 -1.45 -1.40 -1.350.525

0.530

0.535

0.540

0.545

0.550

0.555

0.560

Fig. No. I part magnification in Fig.

0/ xstrainrate。

X/(H2/2

)

β ＝60

β ＝200

β ＝130

45

015.0||

x

015.0||

x

4

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• Results of strain rate

For the same relative streamline, at the points with the same relative horizontal ordinate, calculations showed that the ratio of tensile and shear strain rate to the stagnation point strain rate is very similar for colliding

angles in the range 6–20º. in detail:(the paper)

H.H. Yan, X.J. Li.

Strain rate distribution near welding interface for different collision angles in explosiveWelding

International Journal of Impact and Engineering.2008 ， 35 ： 3-9.

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Solving stain (along stream line)

Fi g. Di agram of streaml i ne (trace)

nn

On

θ τ n y

x o

ξ

Ψ 1 Δ

θ

nn+1

On+1

τ n+1

When the continuums deform, displacement, eddy and distortion of each infinitesimal among them will change. Its orientation and shape will change at any time. According to the steady assumption, the streamline is same as the trace; that is to say, change of strain along the streamline is same as that along the trace. To calculate the strain distribution, the eddy factor must be eliminated.

6

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Calculating Process

：

+ +nn

nTn

ttt tt

EEt

EE

nn

1

1

0 1

limlim tnnTn EEE

1 tEEE nnn

1

jlikklnijn mmEE )()(

jlikklnijn EE )()(

)2,1,,,( lkji

22nn vu

lt

To calculate the natural strain by eliminating the eddy factor

In detail

Transformation relation

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Strain results (For example)

-10 -8 -6 -4 -2 0 2 4 6 8 10-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

Fig. Diagram of strain with different angles at ~ ＝ -1.0

X/(H2/2)

β ＝60

β ＝200

β ＝130

Stra

in

E11

E12

-10 -8 -6 -4 -2 0 2 4 6 8 10-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

Fig. Diagram of strain with different angles at ~ ＝ -2.0

X/(H2/2)

β ＝60

β ＝200

β ＝130

Stra

in

E11

E12

5

78

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Discussion

• The model adopted is simple and ideal• If the viscous-elastic constitutive equations were used to

analyze stain field, explosive welding mechanics will be explained very well.

• If the geometric non-linearity was considered, the Green’s and Almansi’s strain can be used in the future.

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THANK YOU