PSI - Issue 64

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

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2.2. Solution strategy Solution for the un-cracked beam

The load response for the un-cracked phase is constructed considering a set of values of the slip s E at the end of the reinforcement. The slip s ( z ), axial force N ( z ), and rotation ϕ ( z ) profiles associated with a generic value of the slip s E can be determined by solving Eq. (5) with the boundary conditions: ( ) 0 0 s = ( ) 0 f N z = ( ) 0 0 ϕ = ( ) f E s z s = (9) E s are then associated with any value of s E . The corresponding deflection ϕ function. The un-cracked stage ends when the maximum concrete tensile strain attains the concrete cracking strain ε t . Because of symmetry, during the un-cracked stage the maximum concrete tensile strain occurs at midspan, therefore the cracking condition is ( ) ( ) ( ) ,max 0 0 t n t H d ε =χ − =ε , where the curvature at midspan ( ) 0 χ can be evaluated with the third equations in Eq. (5) and the distance from the extrados of the neutral axis at midspan d n (0) is evaluated with Eq. (6). The profiles ( ) E s s z , ( ) E s N z , and ( ) ϕ E s z corresponding to the un-cracked stage at the attainement of ε t are shown in Fig. 3c. These profiles were obtained by assuming a concrete linear elastic behavior and the shape of the CML shown in Fig. 3b, which is described by the equation proposed by Dai et al. (2006). Solution for the beam with one crack at midspan The load response of the beam with one crack at midspan is constructed by considering a set of values of the slip s M at the right face of the midspan crack. Due to symmetry, at midspan the rotation is ( ) ( ) 1 0 2 M c s H d ∆ϕ =− − , where d c 1 is the position of the CRR at midspan. For any ( ) 0 ∆ϕ , Eq. (7) is employed to determine the axial force at midspan ( ) 0 M N s , λ and ( ) 1 c M d s , λ associated with a generic value of λ . Eq. (6) is then solved with the boundary conditions ( ) E s = λ E s v v by integrating the ( ) M s z slip s M at mid-span. The corresponding deflection function. The one-crack stage ends when the maximum concrete tensile strain attains ε t at a certain cross-section z c 2 . The profiles ( ) M s s z , ( ) sM N z , and ( ) M s z ϕ corresponding to ε t for the beam with a crack at midspan are shown in Fig. 3d. Solution for the beam with n cracks The load response of the beam with n cracks is constructed considering a set of values of the slip s M at the right face of the midspan crack. For any s M , Eq (5) is solved for each segment of the beam. Referring to Fig. 4, z cj , ∆ϕ j , and d cj ( j =1,… , n ) are the coordinates of the cracked cross-sections, corresponding relative rotations, and positions of the CRR. The segment between cracks at z cj and z cj+1 will be named block j ( j =1,…, n -1). Block n is the segment between the crack at z cn and z f . For block j , the solution of Eq. (5) consists of the slip, axial force, and rotation profiles, denoted as s ( j ) ( z ), N ( j ) ( z ), and ϕ ( j ) ( z ). For the block 1, the first three boundary conditions in Eq. (10) can be enforced at midspan, whereas the last boundary condition in Eq. (10) cannot be enforced due to the discontinuities at z cj . These three conditions are not sufficient to determine the slip, axial force, and rotation profiles and λ . A fourth condition has to be enforced at z c 2 such that N ( z f )=0. This condition is unknown, since Eq. (7) does not allow to determine ∆ϕ , d f , and N for an assigned λ . An attempt value of d c 2 , named 2 d , is considered and the corresponding ∆ϕ and N , named 2 λ ∆ϕ and 2 N λ , respectively, are determined as a function of λ with Eq. (7). Then, the slip, axial force, and rotation profiles for block 1, named ( ) ( ) 1 s z , ( ) ( ) 1 N z , and ( ) ( ) 1 z ϕ , and λ can be determined by adding the boundary condition ( ) 2 2 c N z N λ = (11) ( ) M s M s v v = λ can be obtained by integrating the M s z ϕ for increasing value of the slip - s E . A profile of slip ( ) E s s z , axial force ( ) E s N z , and rotation ( ) ϕ E s z , and a load can be obtained multiplier λ ( ) 0 M s s = ( ) 0 M N N s , = λ ( 0 ) ( ) 0 ϕ = ( ) 0 ( s H d s , − − M M ) 1 2 c ∆ϕ = λ ( ) 0 f N z = (10) to associate the slip ( ) M s s z , axial force ( ) M s N z , and rotation ( ) M s z ϕ profiles, and a load multiplier M s ( ) λ with any

to the first three boundary conditions in Eq. (10) for the solution of Eq. (5).