Issue 38

S. Bennati et alii, Frattura ed Integrità Strutturale, 38 (2016) 377-391; DOI: 10.3221/IGF-ESIS.38.47 390 f Q f P b Q b P dw dw dw dw d ds ds k ds ds ds ds ds ds ds , , , , * * * *            (A3) Then, we differentiate also Eq. (7) with respect to s and obtain the following expression, whose numerical values turn out to be approximately equal to 1 for current geometric and material properties: b s b s b f f h P E A E I ds ds P E A 2 1 1 4 * 1 1 1      (A4) By substituting Eqs. (4), (5), and (A1) into (A3), and considering (A4), we obtain b b Q f Q s b s b f f h d k P ds E A E I E A 2 , , 1 1 4                          (A5) By further differentiating Eq. (A5) and recalling (A2), we have f Q f Q b Q b Q b Q b s b s b f f d dN d dN dM h d k k ds ds E A ds E I ds E A ds ds 2 , , , , , 2 1 1 2                         (A6) Lastly, by substituting Eqs. (10) into (A6), after simplification, we obtain the following differential equation for the interfacial shear stress: b b f b Q s b s b f f s b h h d kb s k V s E A E I E A E I ds 2 2 , 2 1 1 ( ) ( ) 4 2                (A7) where the shear force is b Q V s q l s , ( ) ( ).   (A8) Solution of the differential problem The differential problem stated in the previous section can be solved analytically. In fact, the general solution to Eq. (A7) for the interfacial shear stress is s C s C s q l s 1 2 ( ) exp( ) exp( ) ( )          (A9) where C 1 and C 2 are integration constants and the constant parameters b b b f s b s b f f s b f b b s b b f f h h h k kb E A E I E A E I b I h E I A E A 2 2 2 1 1 1 and = = 4 2 2 2                       (A10) are also introduced. By imposing the boundary conditions, l (0) 0 ( ) 0        (A11)