PSI - Issue 47

Victor Rizov / Procedia Structural Integrity 47 (2023) 3–12 Author name / Structural Integrity Procedia 00 (2019) 000–000

6 4

where 1 A and  are, respectively, the area and the stress in the crack arm 1, RQ A and RQ  are, respectively, the area and the stress in crack arms 2, 3 and 4, z is the vertical centric axis. Beams of high length to diameter ratio are under consideration in the present paper. Therefore, the distribution of strains is treated by applying the Bernoulli’s hypothesis for plane sections. Hence, the distributions of strains are written as   n DD D D z z   1 2 1 2   , z RQ DD 1 2    , (9) where 1 2 DD  and n z are, respectively, the distribution of strains and the coordinate of the neutral axis of crack arm 1, RQ  is the distribution of strains in crack arms 2, 3 and 4. It should be noted that the neutral axis of crack arms 2, 3 and 4 coincides with y -axis since the axial force in crack arms 2, 3 and 4 is zero. The stresses,  and RQ  , are found, respectively, by substituting of 1 2 DD  and RQ  in (1). Then, by substituting of (1), (2) and (9) in (8), one obtains F R q R R qR H z B R p D D n C              2 4 2 1 2 4 1 2 1 4 3 2 1 1 1 1 2 , (10)        

   

   

1

p



 R R R

z





3 

  

   

  

  3 4 4 R R R R  1

   3 1

2 4 1 4

q



D D

1 2 D D n

D D

4 R B p C 1

1

2

B pe C



1 2

1 2

H

H

H

R R R  2

3



   

  3



4 1 4

q





D D

p p R R R R    1 3

3

B e C

M

  



.

(11)

1 2

1

4

4

1

3

3

H

Equations (10) and (11) are solved with respect to n z and



by using the MatLab computer program. Then,

1 2 DD

0  z in the first formula of (9).



the strain,

, is obtained by substituting of

1 2 CDD



 , are obtained from the following equations for equilibrium in

The strain,

, and the curvature,

2 3 CD D

2 3 D D

2 3 D D :

beam portion,





 LN A

 ST dA F  ,

zdA

 zdA M







,

(12)

ST

LN

( ) 2 A

( ) 2 A

(

)

where 2 A and ST  are, respectively, the area and the stress in the crack arm 2, LN A and LN  are, respectively, the area and the stress in crack arms 3 and 4 in beam portion, 2 3 D D . The strains are written as   2 3 2 3 2 3 nD D D D D D z z     , z LN D D 2 3    , (13) where 2 3 D D  and 2 3 nD D z are, respectively, the distribution of strains and the coordinate of the neutral axis of portion, 2 3 D D , of crack 2, LN  is the distribution of strains in crack arms 3 and 4 in beam portion, 2 3 D D . After substituting of (1), (2) and (13) in (12), the two equations are solved with respect to 2 3 D D  and 2 3 nD D z by the MatLab computer program. The strain, 2 3 CD D  , is found by substituting of 0  z in the first formula of (13). The following equations for equilibrium in the beam portion, 3 4 DD , are used to determine the quantities,





and

:

3 4 CD D

3 4 D D

 ( ) 3 A



 EG A

 BP dA F  ,

zdA

 zdA M







,

(14)

BP

EG

( ) 3 A

(

)

where 3 A and BP  are, respectively, the area and the stress in the crack arm 3, EG A and EG  are, respectively,