PSI - Issue 72

Victor Rizov / Procedia Structural Integrity 72 (2025) 113–119

116

v t y 

y    

(11) (12) (13) (14) (15) (16)

,

v t z 

,

z

1 1 y D D C        , 1 1 1 2 3 z y z

v t C C 1 1   ,

v t y 1  v t z 1 

y 1     1 z

,

,

 2 2 1 b b

  

  

 



 

(17)

.

2 3

1

1

CD D

C

y

Formula (8) expresses the bar axial displacement as a function of the strains, ε C and ε CD 1 D 3 , in the bar portion, D 1 D 2 , and in the left crack arm. Formulas (10) and (13) present the distribution of strains in portion, D 1 D 2 , of the bar and in the left crack arm. The quantities, ε C , κ v and κ z , are the strain in the centre and the curvatures of the bar in the bar portion, D 1 D 2 . The strain in the centre and the curvatures of the left crack arm are ε C 1 , κ v 1 and κ z 1 . Then, the equilibrium of the elementary forces in the cross-sections of portion, D 1 D 2 , of the bar and the left crack arm is considered via the following formulas.   ( ) A y zdA M  , (18)   ( ) A z M ydA  , (19)   ( ) 2 3 1 1 1 A D D y z dA M  , (20)

where M y , M z and M y 1 are the bending moments. They are zero for the bar in Fig. 1. Thus,

0  y M , 0  z M , 0 1  y M .

(21) (22) (23)

2 D , of the bar is analysed. This yields the following formulas:

The equilibrium of section,

 ( ) A  ( ) A

 ( ) A

dA

dA







,

(24)

2 3

D D

1 2 2 b b

    dA

  

 ( ) A

y

y dA





 

.

(25)

2 3 1

D D

Then, the quantities, v κv , v κz , ν εC ,ν C 1 , ν κv 1 and ν κz 1 , are obtained from (8), (18), (19), (20), (24) and (25) by the MatLab with taking into account dependences (9) – (17) and (21) – (23). The final step is to determine the SERR by formula (26).