Issue 29
A. Castellano et alii, Frattura ed Integrità Strutturale, 29 (2014) 128-138; DOI: 10.3221/IGF-ESIS.29.12
R, , Z
r, , z
: f X
= x f X
A deformation gradient
of is assumed to be a smooth function with
f
: F X f X . Since the tube is elastic and isotropic, the general form of the strain energy function is given by W W I , II , III F B B B , where 2 2 2 2 2 T 1 1 I tr , II tr tr tr , III det det 2 2 B F F B B F F FF B F B B B (12) are the orthogonal principal invariants of T : B F F ; thus, the Piola stress takes the form T 1 2 3 DW W W I W III 2 2 2 S F F F F BF F B B (13) with We assume that the inner cylinder is kept fixed, whereas the outer is subject to a uniform angular displacement 0 around its axis. On the bases of only tangential displacements are admitted. This leads to the following mixed boundary-value problem: Div = in S 0 F (15) 1 R 0 at R = R = f X X e (17) For the equilibrium problem (15)-(17), we consider the possibility of an azimuthal shear deformation f defined by r R, R , z Z (18) where , the angular displacement field, is assumed to be a smooth function satisfying 1 2 R 0, R 0 1 on (19) Consequently, in view of (18) we have 2 2 R r r R R , R R I e I F e B e e e e e e (20) where the prime denotes differentiation with respect to R and r z , , e e e is the deformed cylindrical orthonormal basis at r, , z . Notice that the azimuthal shear is an isochoric deformation whose principal invariants (12) are given by 2 2 , I II 3 R II 1 I B B B (21) Moreover, (18) trivially satisfies the displacement boundary conditions (16)-(17) 1 and, in view of (13) and (20), it is easily seen that also the traction boundary condition (17) 2 holds. It remains to study the equilibrium field Eq. (15). The explicitly evaluation of (15) by means of (13) and (20) yields an over-determined system of two ODE for the single unknown function R . In particular, here we consider the class C 2 Levinson-Burgess strain energy function 1 W : 2 3 W W W , W : , W : I II III B B B (14) f X X e f X X e 2 R cos 1 at R = R 0 at R = R 2 1 Z 1 2 R sin at R = R = 0 at R = R R 2 1 2 , = on (16) f X X e S e e 0 F 2 = 0, = on
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