PSI - Issue 47

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

7

5

3 4 DD . The strains are expressed as

the area and the stress in crack arm 4 in the beam portion,



 3 4

D D D D z z    

z EG D D 3 4 

  ,

,

(15)

nD D

3 4

3 4

3 4 nD D z are, respectively, the distribution of strains and the coordinate of the neutral axis in



where

and

3 4 D D

portion, 3 4 DD , of crack 3, EG  is the distribution of strains in crack arm 4 in beam portion, 3 4 DD . After substituting of (1), (2) and (15) in (14), the two equations are solved with respect to 3 4 nD D z and 3 4 D D  by the MatLab computer program. The strain, 3 4 CD D  , is calculated by substituting of 0  z in the first formula of (15). Finally, CUN  and UN  are obtained from the equations for equilibrium of the elementary forces in the cross section of the beam in the un-cracked portion   ) ( UN A UN dA F  ,   ) ( UN A UN zdA M  , (16) where UN A and UN  are the area and the stress in the un-cracked beam portion. The strains, UN  , in the un cracked beam portion are expressed as   nUN UN UN z z     , (17) where nUN z is the coordinate of the neutral axis. After substituting of (1), (2) and (17) in (16), the MatLab computer program is used to solve the two equations with respect to UN  and nUN z . The strain, CUN  , is found by substituting of 0  z in (17). The strain energy is written as UN DD D D D D U U U U U     3 4 2 3 1 2 . (18) where 1 2 DD U , 2 3 D D U , 3 4 D D U and UN U are the strain energies cumulated in beam portions, 1 2 DD , 2 3 D D , 3 4 DD , and in the un-cracked beam portion, respectively. The strain energy in beam portion, 1 2 DD , is expressed as U a u dA a u dA RQ A A DD RQ 0 ) ( 1 01 ( ) 1 1 1 2     , (19) where 01 u and RQ u 0 are the strain energy densities in crack arm 1 and in crack arms 2, 3 and 4, respectively. In principle, the strain energy density is equal to the area enclosed by the stress-strain curve. Thus, by integrating of stress-strain relation (1) and replacing of  with 1 2 DD  , the strain energy density in crack arm 1 is obtained as         1 1 01 1 2 BH BH u B p D D D D . (20)

  

     p

  

  

1

1

p

H



1 2

RQ u 0 . For this purpose,

 is found by the first formula of (9). Formula (20) is used also to calculate

where

1 2 DD



RQ  . In beam portion,

2 3 D D , the strain energy is expressed as

is replaced with

1 2 DD

  

   ,







 LN A

U

a a

 u dA u dA

 

(21)

2

1

0

0

D D

ST

LN

2 3

( ) 2 A

(

)



ST u 0 and

LN u 0 , are obtained by (20). For this purpose,

where the strain energy densities,

is replaced with

1 2 DD



3 4 DD , is written as

LN  . The strain energy in the beam portion,

and

2 3 D D