Issue 29

G. Maurelli et alii, Frattura ed Integrità Strutturale, 29 (2014) 351-363; DOI: 10.3221/IGF-ESIS.29.31 353   1 ( ) 2 s T v v v v        (2) whereas the dynamic equilibrium reads div  v g         (3) Truly-mixed variational formulation By observing that the total stress  may be additively decomposed as 0 1      , the continuous variational formulation of the problem at hand may be obtained by eliminating the strain tensor  in Eq. (1) and (2) testing the resulting equation by two virtual stress fields 0 1 ,   and the equilibrium Eq. (3) by a virtual velocity field w so as to write: find 2 0 1 ( , , ) (div, ) (div, ) ( ) v H H L         such that 0 0 0 0 0 0 0 1 1 1 1 0 1 , , , div  0 , , div  0 , div , div , , E V E A A v A v v w w w g w                                                   (4) 2 0 1 (div, ), (div, ), ( ). H H w L            In more compact form Eq. (4) may be rewritten as . 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 T E V T E A A B A B M v B B v g                                                           (5) It should be emphasized that static and dynamic models differ only in that the mass matrix M may be neglected or considered in statics or dynamics, respectively. Eq. (5) amounts to a system of ordinary algebraic-differential equations (ADE) in the former case, of differential equations in the latter. Imposing Dirichlet and Neumann boundary conditions Within the adopted truly mixed formulations, Dirichlet boundary conditions on the velocity field , on u v v   (6) are imposed weakly thanks to the line integral   u n vds      (7) that appears on the right hand side of Eq. (4 ) 1 and (4) 2 , and in fact, taking no action on a boundary amounts to imposing a homogeneous condition 0, on u v   . Conversely, Neumann traction boundary conditions n t    are to be imposed strongly. However, when an additive stress decomposition is adopted, as is the case herein, i.e. 0 1      , neither 0  nor 1  are known but their sum. Therefore a Lagrange multiplier approach is adopted to enforce weakly a condition of type 0 1 ( ) n t      . Therefore, the resulting linear algebraic-differential system takes the form . 0 0 0 2 0 1 1 1 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 T T E V T T E A A B B A B B v v g M B B t B B                                                                           (8)