Issue34

T. Itoh et alii, Frattura ed Integrità Strutturale, 34 (2015) 487-497; DOI: 10.3221/IGF-ESIS.34.54 492 Fig. 7 shows the shape and dimensions of hollow cylinder specimen employed which has a 12 mm inner diameter, a 14 mm outer diameter, and an 8.5 mm parallel at gauge part. In the figure, a coordinate employed are indicated with principal stresses. σ 1 is the maximum principal stress, σ 2 is the middle principal stress. In this test, σ 1 and σ 2 are equivalent to axial stress σ z and hoop stress σ θ equated by the following equations,       2 1 2 2 2 2 4 4 4 i z o i o i            P D F D D D D (7) 2 i o i        P D D D (8) where F and P are axial load and inner pressure, respectively. D i and D o are inner and outer diameters at gauge part of the specimen. Figure 7 : Shape and dimensions of test specimens (mm) with coordinate. Stress controlled low cycle fatigue tests with proportional zero-peak loading were conducted under 4 types of stress paths shown in Fig. 8. In the figure, each stress path takes the principal stress ratio, λ=0, 0.4, 0.5 and 1.0. The dashed line shows a constant Mises’ equivalent stress at 800 MPa. In the test at  =0.4, peak stress was set at 858 MPa. The other  ’s test at 800 MPa. Middle principal stress σ 2 Maximum principal stress σ 1 (λ=0) (λ=1.0) (λ=0.5) (λ=0.4) Mises stress σ eq =800MPa O Figure 8 : Principal stress path. Fig. 9 shows strain waveforms of axial load ( F ), inner pressure ( P ) and axial stress (  z ), hoop stress (   ). Test frequencies in the uniaxial fatigue test (λ=0) and the biaxial fatigue tests (λ=0.4, 0.5, 1.0) are 0.4 Hz and 0.2 Hz, respectively. Number of cycles to failure (failure life), N f , was determined as the cycle at which the maximum inner pressure was reduced due to leak of oil by initiation of through crack or rupture of the specimen. Tab. 1 summarizes test conditions conducted. In the table,  z max and   max are maximum values of  z and   obtained at the peak load, respectively.  eq max a maximum value of Mises’ equivalent stress (  eq ) equated by r θ z σ θ =σ 2 σ z =σ 1 σ r =σ 3 =0