Issue 46

V. Rizov, Frattura ed Integrità Strutturale, 46 (2018) 158-177; DOI: 10.3221/IGF-ESIS.46.16 164 1 1 i i B i i D i i g g r r g r r      (26) where 1 i i r r r    (27) The radiuses, i r and 1 i r  , are shown in Fig. 2. In (25) and (26), i B f , i D f , i B g and i D g are material properties ( i D f and i D f govern the gradient of i f and i g , respectively). The distribution of shear strains in radial direction is written as m b r r    (28) By substituting of (24), (25), (26) and (28) in (23), one derives     1 4 4 5 1 1 1 2 1 1 5 4 5 i n m i i i i i i i b T r r r r r                    (29) where 1 i i    (30)   2 2 2 1 1 i i i B i B m i D i i i i i b i f g g r r r                 (31) i D i i f    (32) 1 i B i i i f r     (33) It should be mentioned that at 0 i B g  and 0 i D f  Eqn. (29) transforms in   1 4 4 1 1 1 2 i i n m i i i b B T r r r f         (34) The fact that (34) is exact match of the equation for equilibrium of a multilayered circular shaft made by linear-elastic homogeneous layers loaded in torsion [14] is an indication for consistency of (34) since at 0 i B g  and 0 i D f  the non- linear stress-strain relation (24) transforms in the Hooke’s law assuming that 1/ i B f is the shear modulus in the i -th layer. Eqn. (29) should be solved with respect to m  by using the MatLab computer program. Eqn. (29) is used also to determine the shear strain at the periphery of the cross-section of the un-cracked shaft portion. For this purpose, 1 n , b r and m  are replaced, respectively, with n , R and q  in (29) and (31).