PSI - Issue 28

Victor Rizov et al. / Procedia Structural Integrity 28 (2020) 1237–1248 Author name / Structural Integrity Procedia 00 (2019) 000–000

1241

5

  1 A

N

dA  ,

(13)

1

where 1 A is the area of the internal crack arm cross-section, 1 N is the axial force in the internal crack arm. By combining of (8) and (13), one derives

3 1

4 1

  

  

 2 2 1 N R q q R q R  .   2 3 1 2 1 1 2 1 1

(14)

Equations (10), (11), (12) and (14) should be solved with respect to  , 1 q , 2 q and 3 q by using the MatLab computer program for particular material properties, geometry and loading of the bar. After that, * 01 u is obtained by substituting of (3) and (8) in (5). Formula (5) is applied also to calculate the complementary strain energy density in the cross-section of the external crack arm behind the crack front by replacing of  with f  where f  is the stress in the external crack arm. However, since in the external crack arm (Fig. 1) the radius varies in the interval, 2 1 R R R   , the stress, f  , is expanded in series of Taylor. The first three members are kept. Therefore, the series is written as     2 3 2 1 ( ) a f a f f f R q q R R q R R       , (15)

where

2 1 R R R a  

.

(16)

2

1 f q , 2 f q and 3 f q , are determined in the following way. First, by substituting of (15) in the

The unknown coefficients,

Ramberg-Osgood equation (2), one obtains





n 1









2

2

   E q q R R q R R a f a f f 3 2 1 ( ) ( 

)

n H q q R R q R R 1 3 2 1     f a f f

a

f 





,

(17)

a R R  formula (17) transforms in

where f  is the longitudinal strain in the external crack arm. At

n f 1

H q

  E q f 1

1 .

f 

(18)

n 1

At a R R  the first and second derivatives of (17) with respect to R take the forms