Issue 43

L.C.H. Ricardo, Frattura ed Integrità Strutturale, 43 (2018) 57-78; DOI: 10.3221/IGF-ESIS.43.04

  2 0 r



ys P B

dx

(2.6)

ys

where B=thickness λ = 1

for plane stress

λ = 3 for plane Irwin’s yielding factor for plane strain [7]

For equilibrium conditions, the force balance 

    0 s ys P P

leads to the determination of the of the plastic zone size

Hence,

r

r

1

2



    ys

 0

 0







dx

dx

0

(2.7)

ys

Considering:



   I







 0 lim 2 yy r 



 











  , 0

K

r f

r for

r

2

(2.8)

yy

yy

yy

1

Inserting (2.2) into (2.8) and integrating yields

r

r

1

2

K

  

  

 0

 0

I











dx

dx

0

(2.9)

ys

ys



r

2

1 r K

2





  I r



 

r r

(2.10)

0

ys

1 2

2











 

r

r r

(2.11)

2

0

y

ys

1

1 2

The elastic stress can be defined by



 y



(2.12)

ys

Inserting eq. (2.12.) into (12.3) gives 2r 1

= r 1

+r 2

which implies that r 1

=r 2

and from Fig. 2, r 1

= r 1

+r 2.

Hence, a e

= a+r is the

virtual crack length proposed by Irwin [6]. Obviously, eq. (2.14) provides the effective stress intensity factor

 I

  a

K

(2.13)







 

 

  a

K

a r

(2.14)

I

e

The plastic zone size can be calculated by eqs. 2.4 and 2.5. This K I finite specimen size and plasticity. Now, inserting eqs. (2.4) into (2.14) yields.

equation is the corrected stress intensity factor due to

           2 2 ys a r

(2.15)

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