PSI - Issue 64

Francesco Focacci et al. / Procedia Structural Integrity 64 (2024) 1557–1564 Author name / Structural Integrity Procedia 00 (2019) 000–000

1563

7

d cj +1 ∆ϕ j +1

d cn ∆ϕ n

d c 2 ∆ϕ 2

d cj ∆ϕ j

z

,…,

,…,

j

z f

n

1

H

z c 1 = 0

z c 2

z cj

z cj+1

z cn

Fig. 4. Position of cracks.

It should be noted that the solution of Eq. (5) with the first three boundary conditions in Eq. (10) and the condition in Eq. (11) depends on 2 d and includes the value of λ corresponding to the actual value of s M . The solution of Eq. (5) for block 2 requires only three boundary conditions since λ is known from the solution corresponding to block 1. For j =2, 3,… n , the slip, axial force, and rotation at the beginning of block j ( cj z z = ) are known from the solution corresponding to block j -1. Thus, the slip ( ) ( ) j s z , axial force ( ) ( ) j N z , and rotation ( ) ( ) j z ϕ profiles for block j can be obtained by solving Eq. (5) with the boundary conditions ( ) ( ) ( ) ( ) 1 j j cj cj j s z s z H d − λ = −∆ϕ − ( ) ( ) ( ) 1 j cj cj N z N z − = ( ) ( ) ( ) 1 j j cj cj z z − λ ϕ =ϕ +∆ϕ (12) j d are determined with Eq. (7) based on λ determined for the first block and the axial force . This procedure continues up to block n . The slip ( ) ( ) n s z axial force ( ) ( ) n N z , and ( ) ( ) n z ϕ profiles for ( ) ( ) 0 n f N z = , which is required at the end of the reinforcement. This is because the described procedure starts from the attempt value 2 d at the cross-section z c 2 . Therefore, a value of ( ) ( ) n f N z can be associated with any value of 2 d and the function ( ) ( ) ( ) 2 n E f N d N z = can be defined, which associates 2 d with the axial force at the end of the reinforcement. The correct position d c2 can be determined at any load step (i.e., for any s M ) by enforcing ( ) 2 0 E c N d = . The described procedure can be then applied with the correct value d c2 to determine the profiles of slip, axial force, and rotation, and λ associated with any s M . This procedure, which involves the preliminary solution of one equation in one unknown, i.e., ( ) 2 0 E c N d = , is not advantageous from the numerical point of view since the boundary conditions for each block are enforced at its initial cross-section instead of at both cross-sections delimiting it. For this reason, a slightly different procedure was implemented, which is more time consuming but more reliable from the numerical point of view. The procedure starts from an array of guess positions of the CRR at the end of the blocks { } 1 2 1 n d ,d ,...,d − = d     , where j d  is the guess value of the position of the CRR at the end of block j (i.e., at crack j +1). The relative rotation  j ∆ϕ and the axial force cj N  at the right end of the j th block corresponding to the guess value j d  can be determined as a function of λ using Eq. (8) with c j d d =  . Then, the slip, axial force, and rotation profiles corresponding to the block 1, named ( ) ( ) 1 s z  , ( ) ( ) 1 N z  , and ( ) ( ) 1 z ϕ  , and λ can be determined by adding condition ( ) 2 1 c c N z N =  to the first three of the boundary conditions in Eq. (10). The slip ( ) ( ) j s z  , axial force ( ) ( ) j N z  , and rotation ( ) ( ) j z ϕ  profiles corresponding to block j are determined with Eq. (5) with the boundary conditions ( ) ( ) ( )  ( ) 1 1 1 j j cj cj j s z s z H d − − − = −∆ϕ −   ( ) ( ) ( )  1 1 j j cj cj z z − − ϕ =ϕ +∆ϕ  ( ) 1 cj cj N z N + =  (13) The slip ( ) ( ) n s z  , axial force ( ) ( ) n N z  , and rotation ( ) ( ) n z ϕ  profiles corresponding to block n (between the crack at z cn and the end of the reinforcement) are determined with Eq. (5) with the boundary conditions ( ) ( ) ( )  ( ) 1 1 1 n n cn cn n s z s z H d − − − = −∆ϕ −   ( ) ( ) ( )  1 1 n n cn cn z z − − ϕ =ϕ +∆ϕ  ( ) 0 f N z = (14) ( ) ( ) 1 j cj N z − this block are obtained by solving Eq. (5) with the boundary conditions in Eq. (12) with j = n . The ( ) ( ) n N z profile obtained does not necessarily satisfy the condition where j λ ∆ϕ and

The axial force profiles obtained for the n blocks do not necessarily satisfy the n -1 conditions ( ) ( ) ( ) ( ) 1 0 j j j cj cj N N z N z − ∆ = − =  ( 2 3 j , ,...n = ) that are required by the continuity of the axial force. This is because