PSI - Issue 54

Victor Rizov et al. / Procedia Structural Integrity 54 (2024) 468–474 Victor Rizov/ Structural Integrity Procedia 00 (2019) 000 – 000

472

5

The distributions of properties in the width direction are described by the following exponential laws

b 2

b 2

b 2

b 2

y

y

y



y







f E E e 

f

f E E e 

f

1

b    e 1 3

4

e    1 2





b

b

b

,

,

,

,

(12)

1

1

2

2

2

1





b 2

b 2

b 2

y

y

y







y b    ,

2 b

f

f

f

e    3 5

e    4 6

b    S e 7







b

b

,

,

,

(13)

3

4

S

2

where 1 f , 2 f , 3 f , 4 f , 5 f , 6 f and 7 f are parameters. The SERR,

i i G dU dA / * 

i G , are found for each longitudinal crack. For this purpose, formula

,

m i A is the i -th crack area, m is the crack number. For control of the solutions, the energy balance is examined which yields   i i i h dU da G M h d da / / ) (1/ ) ( / )(    where U and  are the strain energy and the rotation of the leftmost crack arm. The results yielded by the two solutions are the same. 3. Numerical results From view point of practical engineering, it is useful to examine the effects of such factors as continuous material inhomogeneity, time and viscoplastic behaviour on the SERR for a beam with three longitudinal cracks. 0.01  b m, 0.012  h m and 0.40  l m are used in the calculations. i 1,2,...,  is used where * U is the complementary strain energy,

1.2

1.8

0.6

Fig. 4. Variation of the SERR with 1 f (curve 1 – at

2  f

2  f

2  f

, curve 2 – at

and curve 3 – at

).

How the SERR varies with time for crack 1 can be seen in Fig. 3. Unit ( ) II of the rheological model starts to deform at 1 t t  . This is the cause for the growth of the SERR at 1 t t  . The parameters, 1 f and 2 f , cause a variation of the SERR as depicted in Fig. 4. One can observe that the SERR lessens when 1 f and 2 f increase.