Issue 41

V. Rizov, Frattura ed Integrità Strutturale, 41 (2017) 491-503; DOI: 10.3221/IGF-ESIS.41.61 497 Formula (21) can also be applied to obtain the distribution of complimentary strain energy density, * 0 U u , in the un-cracked beam portion. For this purpose, r , 1 y , 1 z , 1 1 n z and 1  have to be replaced with 0 , 2 y , 2 z , 2 2 n z and 2  , respectively ( 2 2 n z is the neutral axis coordinate of the cross-section of un-cracked beam portion, 2  is the curvature of un-cracked beam portion). The quantities, 2 2 n z and 2  , can be determined from Eqs. (18) and (19). For this purpose, r , 1 1 n z , 1 h and 1  have to be replaced with 0 , 2 2 n z , 2 h and 2  , respectively. Finally, by substituting of * 0 L u , * 0 U u and (8) in (7), one derives         2 1 1 1 0 1 2 1 1 2 2 1 1 3 1 1 m m m u u u u u B b B b B br m G b m m m h m                               1 2 2 1 1 2 1 1 2 1 2 2 2 1 2 1 n m m m m u u u u u u z B br m m h                      3 3 2 2 1 2 1 1 2 1 2 2 2 1 3 2 f q f q f q f q u u u u u u u u n q q q q u u u u u u u u u z B bq f q f q h                                     2 1 1 1 2 f q f q u u u u n q q u u u z f q                     (22)       1 1 1 0 1 2 1 2 1 1 3 1 m m m u u u u u u B b B b m b m m m                       3 3 2 2 2 2 2 1 2 1 2 2 2 1 3 2 f q f q f q f q u u u u u u u u n q q q q u u u u u u u u u u u u u z B bq f q f q h                                     2 2 2 1 2 f q f q u u u u n q q u u u u u z f q                     where 1 u m m   , / u u u f q m  ( u f and u q are positive integers), 1 2 2 u n h z    and 2 2 2 u n h z     . Formula (22) calculates the strain energy release rate in the beam configuration shown in Fig. 1 when the beam mechanical behavior and the material gradient are described by formulae (1) and (2), respectively. It should be noted that at m =1, 0 B E  , 1 2 0 B B   and 1 h h  formula (22) yields 2 2 3 21 4 y M G Eb h  (23) which is exact match of the expression for the strain energy release rate when the beam considered is linear-elastic and homogeneous [26]. In order to verify (22), the fracture is analyzed also by using the J -integral written as [27] 0 cos x y u v J u p p ds x x                       (24)