Issue 42

R. Pawliczek et alii, Frattura ed Integrità Strutturale, 42 (2017) 30-39; DOI: 10.3221/IGF-ESIS.42.04 34 1. Input data: ε max ε min ε a ε m 2. Calculation of σ max where K’(ε m ) and n’(ε m ) are defined for actual mean strain value: 1 max max n a E K              3. Calculation of ε Apl : max max Apl E      4. Calculation of σ a where K’(ε m =0) and n’(ε m =0): 1 n a a a E K             5. Calculation of ε apl : a a apl E      6. Calculation of the mean stress σ m : ( ) m m ap Apl E        7. Calculation of the damage accumulation degree for given stresses (σ a , σ m ) and the fatigue life can be predicted. Additionally, in order to define the stress history parameters, some kinematic models of cyclic deformation were used. Usually, those criteria base on the von Mises’s plasticity surface as the yield criterion. For such calculation then the flow rule for a plastic strain increment calculation must be specified. Additionally, during computation with use of the constitutive equations it is important to define the way the plasticity surface is translated. Translation can be described by some hardening rules. During numerical tests it was found that the best results of calculations were obtained for Garud-Mroz [12] and Chu [13] models of the constitutive equations of the cyclic plasticity description. These cyclic plasticity models are used very often for energy based criteria for fatigue life estimation, eg. Lachowicz [14], to describe stress-strain state and to calculate energy dissipated in material. F ATIGUE TESTS The fatigue tests conducted included subjecting the samples made of S355J0 steel to fatigue loads with including the mean load values. The Tab. 1 presents properties of the tested materials. E GPa σ y MPa σ u MPa ν n’ K’ MPa 210 357 535 0.30 0.2074 1323 C Mn Si P S Cr Ni Cu Fe 0.20 1.49 0.33 0.023 0.024 0.01 0.01 0.035 others Table 1 : Strength properties and chemical composition (%) of S355J0 steel. The tests were performed under bending loads applied in blocks with different mean load values. Two paths of the change in the mean load value in blocks were applied (Fig.4): a) mean load was increased from zero to maximum value for the following sequences in blocks of loading, b) mean load was decreased from maximum to zero value in successive blocks of loading. The mean value of the load was as follow: 1: σ m = 0, 2: σ m = ⅓ σ a , 3: σ m = σ a . The whole number of cycles N b in block was divided to three equal parts n 1 =n 2 =n 3 = ⅓ N b . The block applied in the tests had a length N b =15000 cycles.