Issue 53

S. Kirin et alii, Frattura ed Integrità Strutturale, 53 (2020) 345-352; DOI: 10.3221/IGF-ESIS.53.27 349  S r = S n / S c =804/750= 1.07 for p=12.05 MPa, 1.43 for p=12.05 MPa and „1/2“ crack.  K I =1.39(pR/t)  a=1.39·402·  ·11.75= 3402 MPa  mm for p=9.02 MPa and a/t=0.25 or 4536 MPa  mm for p=12.05 MPa and a/t=0.25.  K I =2.71(pR/t)  a=2.71·402·  ·23.5= 9380 MPa  mm for p=9.02 MPa and a/t=0.5 or 12507 MPa  mm for p=12.05 MPa and a/t=0.5.  K Ic =  (EJ Ic /(1- ν 2 )= 8548 MPa  mm (J Ic =316.6 N/mm [13]).  K r = K I / K Ic =3402/8548= 0.40 (p=9.02, ¼) , 4536/8548= 0.53 (p=12.05, 1/4) , 9380/8548= 1.08 (p=9.02 , 1/4), 12507/8548= 1.44 (p=12.05, ½) . Therefore, the FAD coordinates obtained for cracks treated as through thickness cracks with depths being ¼ and ½ half of thickness, are as follows, Fig. 1: For ¼ crack, p=9.02 (0.71, 0.40), p=12.05 (0.95, 0.53). For ½ crack, p=9.02 (1.07, 1.08), p=12.05 (1.43, 1.44). Clearly, this is overconservative estimation since both material and crack geometry are presented in conservative way. Anyhow, even this simple analysis indicates detrimental effect of over-pressurizing, since the points corresponding with it are way out of the safe region (1/2 crack), or very close to the limit curve (1/4 crack), Fig. 3. Less conservative assumption would be to consider cracks as they are, i.e. treat them as surface cracks. In that case reduction of cross-section is negligible, so S n and S r are calculated as follows:  S n =pR/t= 9.02x2.1/0.047=402 MPa,  S n =pR/t=12.05x2.1/0.047=536 MPa.  S r =Sn/Sc=402/750=0.54  S r = S n / S c =538/750=0.72 Stress intensity factor for surface edge crack in a plate can be evaluated using the following procedure from [14]. Plate is taken as better approximation than thin cylinder, since the ratio B/r i is just 47/2100=0.023, i.e. much closer to 0 than to 0.1, which is the lowest value for cylinders in [14]. Geometry parameters F and Q are calculated for tensile plate with a surface crack (a/2c=0.13 for ¼ crack and ≤ 1) at the point of maximum stress intensity factor ( φ =900) as follows: 1.65 1 1.464 1.08 c Q a          for ¼ crack and 1.22 and for ½ crack, 2 4 1 2 3 1.102 w a a F M M M gf f t t                          for ¼ crack and 1.033 and for ½ crack, 1 1.13 0.09 1.106 c M a                for ¼ crack and 1.108 and for ½ crack, 2 0.89 0.54 1.39 1 / M a c     for ¼ crack and 0.7 and for ½ crack, 24 3 1 0.5 14 1 0.589 0.65 a M a c c              for ¼ crack and -0.355 and for ½ crack,   2 3 1 0.08 0.15 1 cos 1 c g t                     1 1 2 4 2 2 2 sin cos 1, sec 1 2 w c c a f f a W t                                       