PSI - Issue 43

V.I. Golubev et al. / Procedia Structural Integrity 43 (2023) 29–34 V.I. Golubev et al. / Structural Integrity Procedia 00 (2022) 000 – 000 For | +1 | < | 3 3 + 1 | 3 +1 = 3 + 1 , +1 = 0, = (10) Here indices + 1 and correspond to current and next time layers, Δ is the time step, index corresponds to the intermediate elastic solution. It is implied, that values of +1 and +1 have been computed previously based on any explicit finite difference scheme. Let ’ s construct the explicit-implicit numerical scheme with the higher order of approximation. Now we will not write explicitly the spatial derivative approximation, to leave it for future discussion (Golubev et al., 2021a; Golubev et al., 2022b). The implicit time approximation of the second order for tangential stresses is: 3 +1 − 3 Δ = ( ,3+1 + 3 , + 1 ) − 2 {〈 |⃗ +1 | | 3 3+1 | − 1〉 3 +1 | ⃗ +1 | + 〈 | ⃗ | | 3 3 | − 1〉 3 | ⃗ | }, (11) 3 +1 = 3 + 1 − 2 1 {〈 |⃗ +1 | | 3 3+1 | − 1〉 3 +1 | ⃗ +1 | + 〈 | ⃗ | | 3 3 | − 1〉 3 | ⃗ | }. (12) This nonlinear system of equations with respect to 3 +1 may be rewritten as: 3 +1 = 3 + 1 − 2 1 {〈 |⃗ +1 | | 3 3+1 | − 1〉 3 +1 | ⃗ +1 | + 〈 | ⃗ | | 3 3 | − 1〉 3 | ⃗ | }, (13) 3 +1 + 1 2 {〈 |⃗ +1 | | 3 3+1 | − 1〉 3 +1 | ⃗ +1 | + 〈 | ⃗ | | 3 3 | − 1〉 3 | ⃗ | } = 3 + 1 . (14) We convolve these equations with 3 +1 , 3 , 3 + 1 . Let’s consider new notations for unknown variables 2 = 3 +1 ∙ 3 +1 , 2 = 3 ∙ 3 +1 , 2 = 3 + 1 ∙ 3 +1 , (15) and for calculated previously variables 2 = 3 ∙ 3 , 2 = 3 ∙ 3 , Σ 2 = 3 ∙ 3 . (16) It is a system of algebraic equations on X, Y, Z : 2 + 1 2 {〈 | 3 3+1 | − 1〉 2 + 〈 | 3 3 | − 1〉 2 } = 2 , (17) 2 + 1 2 {〈 | 3 3+1 | − 1〉 2 + 〈 | 3 3 | − 1〉 2 } = 2 , (18) 2 + 1 2 {〈 | 3 3+1 | − 1〉 2 + 〈 | 3 3 | − 1〉 2 } = Σ 2 . (19) To be short, we use more notations as ΔT = 〈 | 3 3+1 | −1〉 2 , ΔX = δ + 〈 | 3 3+1 | −1〉 2 . (20) So, the previous system will be transformed into 2 ΔX + 2 ΔT = 2 , (21) 2 ΔX + 2 ΔT = 2 , (22) 2 ΔX + 2 ΔT = Σ 2 . (23) The combination of second and third equations leads to 2 = 2 − 2 ΔX ΔT , 2 = Σ 2 − 2 ΔX ΔT . (24) Based on it, the first equation is solved as 2 ΔX = Σ 2 − 2 ΔX ΔT − ΔT S 2 − 2 ΔX ΔT , (25) or 2 ΔX = Δ 2 , Δ 2 = ( Σ 2 − 2 ΔT) − ΔT( S 2 − 2 ΔT) . (26) Δ , | +1 | = √ 3 + 1 3 + 1 , = 1,2 . 31 3