Issue 41

V. Rizov, Frattura ed Integrità Strutturale, 41 (2017) 491-503; DOI: 10.3221/IGF-ESIS.41.61 496       2 1 1 0 1 1 1 2 1 1 1 2 2 1 3 1 1 m m m m m B b B b B br N m m h m                            1 2 2 1 1 2 1 1 1 2 1 2 2 2 1 2 1 m n m m m m z B br m m h                       (18) 3 3 2 2 1 2 1 1 1 2 1 2 2 2 1 3 2 f q f q f q f q m n q q q q z B b q f q f q h                                      2 1 1 1 2 f q f q n q q z f q                     2 2 2 1 1 2 1 0 1 1 2 2 1 1 3 2 m m m m m y B b B br M B b m h                           3 3 1 1 1 1 2 1 1 2 1 2 2 2 1 1 3 f q f q m n q q m m z B br q m f q h                                2 2 2 1 1 1 1 1 2 1 2 2 2 f q f q f q f q n n q q q q z z f q f q                                 (19) 4 4 3 3 1 2 1 1 1 2 1 2 2 3 1 4 3 f q f q f q f q m n q q q q z B b q f q f q h                                     2 2 2 3 1 1 1 1 1 2 1 2 3 2 f q f q f q f q n n q q q q z z f q f q                                  where 1 1 1 1 / 2 n h z    , 2 1 1 1 / 2 n h z     and / f q m  ( f and q are positive integers). In (19), 1 N and 1 y M are determined by (15). It is obvious that at m =1 the non-linear stress-strain relation (1) transforms into the Hooke’s law, assuming that B = E (here E is the modulus of elasticity). This means that at m =1 Eq. (19) should transform into the formula for curvature of linear-elastic beam. Indeed, at m =1, 0 B E  and 1 2 0 B B   Eq. (19) yields 1 1 3 1 12 y M Ebh   (20) which is exact match of the formula for curvature of a linear-elastic homogeneous beam of width, b , and height, 1 h . Eqs. (18), and (19) should be solved with respect to 1 1 n z and 1  by using the MatLab computer program. By substituting of (12) and (16) in (11) the distribution of complementary strain energy density in the lower crack arm is written as     1 1 2 2 1 1 1 1 1 * 1 0 0 1 2 2 2 4 1 m m n L m z z r z y u B B B m b h                  (21)