Issue 41

F. Berto et alii, Frattura ed Integrità Strutturale, 41 (2017) 260-268; DOI: 10.3221/IGF-ESIS.41.35 263                                                                         /2 1 2 2 r, ψ,n ( 2)sin ( 2)sin 4 2 2 2 2 2 cos 2 cos 2 2 n i i j j n n n r a n a n n n a n a n (5b)                                                                          /2 1 2 2 r, ψ,n ( 2)cos ( 2)cos 4 2 2 2 2 2 sin 2 sin 2 2 n r i i j j n n n r a n a n n n a n a n (5c) For n=1 the subscripts are i=1 and j=2; for n=2 we have here only j=3 (because the correspondent terms containing a i disappear). Finally, for n>2, the subscripts are i=2n  2 and j=2n  1. Eqs. 5a-c are given also in [13, 14] where they are splitted into symmetric and skew-symmetric terms. Analogous equations are given also in [15-17] with inclusion of the terms linked to the rigid translation of the slit tip and the rigid body rotation with respect to the same point. By considering the first 13 parameters (a 1 ,…, a 13 ) of the Williams’ series, the stress field can be explicitly derived:                                               2 1 1 2 2 3 1/2 1/2 4 4 5 5 2 6 7 7 3/2 8 8 9 9 ψ ψ 1 3 3 cos ψ 5 cos 3 sin ψ 5 sin 4 cosψ 2 2 2 2 4r ψ ψ r 5 5 9 cos 3 cos ψ 9 sin 15 sin ψ 4 2 2 2 2 8 cosψ sin ψ 2 cosψ 6 cos3ψ r 3 7 3 7 5 cos ψ 15 cos ψ 5 sin ψ 35 sin ψ 4 2 2 2 2 rr a a a a a a a a a r a a a a a a a                  2 10 11 5/2 12 12 13 13 r 6 sin 4ψ 12 cos4ψ r 5 9 5 9 7 cos ψ 35 cos ψ 7 sin ψ 63 sin ψ 4 2 2 2 2 a a a a a a (6a)                                                                     3 2 2 1 2 3 1/2 3 3 1/2 4 4 5 5 3 2 6 7 3/2 8 8 9 ψ ψ ψ 1 cos 3 sin cos 4 sin ψ 2 2 2 r ψ ψ ψ 15 15 5 r 9 cos 6 cosψ cos sin sin ψ 2 2 4 2 4 2 r 8 sin ψ 24 sin ψ cosψ r 3 7 3 35 cos ψ 15 cos ψ 35 sin ψ 4 2 2 2 a a a a a a a a a a a a                               9 2 2 2 2 10 11 11 5/2 12 12 13 13 7 35 sin ψ 2 r 24 sinψ sin 2ψ 24 sin ψ 48 sin ψ cos 2ψ r 5 9 5 9 63 cos ψ 35 cos ψ 63 sin ψ 63 sin ψ 4 2 2 2 2 a a a a a a a a (6b) Along the bisector line the stress components turn out to be: