PSI - Issue 2_B

Tuncay YALÇINKAYA et al. / Procedia Structural Integrity 2 (2016) 1716–1723 Tuncay Yalc¸inkaya and Alan Cocks / Structural Integrity Procedia 00 (2016) 000–000

1722

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2.3. Work of fracture

The work of fracture could be calculated through the following integral in terms of equivalent traction ¯ T and equivalent separation δ e or through normal and tangential components separately

G = � T n d δ n + T t d δ t = � ¯ Td δ e

(13)

∂ ¯ σ ∂ T t

∂ ¯ σ ∂ T n

Note that d δ n = d δ e

and d δ t = d δ e

. Therefore

∂ ¯ σ ∂ T n

∂ ¯ σ ∂ T t �

G = � � T n

d δ e

+ T t

(14)

1 2 where

Remember that ¯ σ = � T 2

2 t / g 2 ( f ) �

n / g 1 ( f ) + T

2

(1 − f ) 2 3

1

1

f �

2 + �

g 1 ( f ) = (1 − f )

and g 2 ( f ) =

(15)

ln

√

3

T 2 n ¯ σ g 1 ( f )

T 2 t ¯ σ g 2 ( f ) � ,

T n ¯ σ g 1 ( f )

T t ¯ σ g 2 ( f )

¯ σ ∂ T t

¯ σ ∂ T n

∂ ¯ σ ∂ T n

∂ ¯ σ ∂ T t �

. The integrand in (14) becomes � T n

= �

and ∂

Then ∂

+ T t

+

=

=

and (14) becomes

f

f

G = �

¯ σ d δ e = �

σ y d δ e

(16)

f 0

f 0

(7) and (12) gives

1 2

d δ e = � d δ 2

n g 1 ( f ) + d δ 2

t g 2 ( f ) �

(17)

d δ t l

2 h a

d δ n h

(1 − f ) + 2 � f

and dh = d δ n −

d δ t

d f =

For proportional loading assume that d δ t / d δ n = α ,

1 2

2 g

d δ e = d δ n � g 1 ( f ) + α

2 ( f ) �

(18)

(1 − f ) h

2 h α l � f

and dh = d δ n    1 −

  

l �

d f = d δ n �

α

2 � f

+