PSI - Issue 44

Laura Giovanna Guidi et al. / Procedia Structural Integrity 44 (2023) 1284–1291 Laura Giovanna Guidi et al. / Structural Integrity Procedia 00 (2022) 000–000

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representation in a dimensionless plane that relies on the aforementioned parameters involved in the definition of critical load, the authors (Brandonisio et al., 2017) (Brandonisio et al., 2019) have proposed a representation of buckling limit, assuming both conditions (11): S γ ; ≤ 1 σ G ≤ σ 0'1#,3 (11) In a dimensionless plan, having on the horizontal axis the ratio γ /S 2 and on the vertical one the dimensionless vertical stress σ v / σ crit , this system of equation defines the area of the stability domain. This type of representation for stability limit deformation has been argued also by AIJ (2016) in the case of lead and high damping rubber bearings. This kind of formulation shows that the buckling load decreases corresponding to the amount of shear deformation d H . So, stable shear deformation can be obtained from the following equation (12): ? ? =5$> =1- , - ! = 1- B A C (12) 3. Code requirements for stability issue To prevent buckling occurrence for rubber devices, the national building code (NTC2018 and the corresponding Circular) assumes a design safety factor for critical load equal to 2. As indicated in (13), this means that the buckling ratio has to be at least equal to 2: V< : ; V 0'1# → < =5$> < ≥ 2 (13) where, V crit is defined in accordance to (14), as function of: G dyn , as the dynamic rubber shear modulus; A r , as the reduced area; S 1 , as the primary shape factor; b min = f for circular bearing; t e as the total rubber thickness. V 0'1# = G dyn · A ' ∙ S : ∙ H I$J # % (14) The withdrawn European Standards EN 15129:2009 (replaced EN 15129:2018) defined the buckling load for rubber devices, having S 1 >5, in accordance to (15). For circular devices, it assumed: a’= f , as the effective diameter, and λ = 1,30. V 0'1# = λ · G · A ' ∙ S : ∙ M KL N (15) To avoid buckling instability, also in this case the applied vertical force, V, has to be lower than half of critical load, V crit (V< 0,5V 0'1# ) In particular, the following conditions (16) (17) should be satisfied: for < =5$> " < V< < =5$> ; → 1- < ;< =5$> ≥ 0 ,7· , - ! (16) for V< < =5$> " → · , - ! ≤ 0,7 (17) According to previous formulations, in the case of V= 0,50V crit (16), the dimensionless horizontal displacement is equal to zero (d H / f = 0), i.e. no shear deformations can occur. When V= 0,25V crit (17), the maximum allowable horizontal displacement corresponds to 70% of device diameter. At the same time, for a horizontal displacement equal to 50% of device diameter (d H / f = 0,50), the maximum compressive force cannot exceed V= 0,325 V crit .