Page 258 - The Combined Finite-Discrete Element Method
P. 258

DISCRETE CRACK MODEL      241

           where the variable D is given by

                                         
                                             0,
                                                   if δ ≤ δ t
                                         
                                             1,     if δ> δ c
                                         
                                     D =                                        (7.13)
                                          δ − δ t
                                                , otherwise
                                         
                                           δ c − δ t
           while the parameters a, b and c are obtained from experimental stress displacement curves
           by curve fitting. Note that for any value of these parameters, the above heuristic formula
           results in a bonding stress of f t for D = 0 and a bonding stress equal to zero for D = 1.
           The tangent at the stress displacement curve at D = 0 is horizontal. Thus, parameters a,
           b and c control the slope of the curve at D = 1 and the shape of the curve (curvature) at
           D = 0 together, with the inflection point.
             In the discrete crack model, it is assumed that the crack walls coincide with the finite
           element edges. Thus initially the total number of nodes for each of the finite element
           meshes (every single discrete element is associated with its separate finite element mesh)
           is doubled, and nodes are held together through a penalty function method. Thus the
           separation δ t is a function of the penalty term p employed. In the limit no separation of
           adjacent edges takes place before stress f t is reached, i.e.

                                            lim δ t = 0                         (7.14)
                                            p→∞
           With increasing separation δ> δ t the bonding stress decreases, and at separation δ> δ c
           it is zero and the crack is assumed to propagate.
             In finite element discretisation of the governing equations, only approximate stress
           and strain fields close to the crack tip are obtained. With the bonding stress model as
           described above, the stress and strain fields close to the crack tip are influenced by the
           magnitude and distribution of the bonding stress close to the crack tip. For the bonding
           stress to have a significant effect on stress distribution results, it is necessary that the size
           of finite elements close to the crack tip be smaller than the actual size of the plastic zone.
           The coarser mesh results in bonding stress in all elements close to the crack walls being
           reduced to zero, except for the few elements adjoining the crack tip. The propagation of
           the crack is therefore influenced by the orientation of those elements close to the crack
           tip. The coarse finite element mesh does not accurately represent the stress field in the
           proximity of the crack tip, and as a result, the stress field obtained is influenced by the
           mesh topology close to the crack tip. i.e. the de-bonding and separation of crack walls
           occurs on an element-by-element basis.
             One way to avoid this problem is to have an element size close to the crack tip much
           smaller than the size of the plastic zone. The approximate length of the plastic zone
             for a plane stress mode I loaded crack can be approximated from Muskhelishvili’s
           exact solution for a crack loaded in mode I. For an infinite body under plane stress
           conditions, Muskhelishvili’s solution gives at points y = 0 normal stress σ y and crack
           opening displacement δ as a function of the coordinate x:

                                        1              4a  #
                                              and δ = σ    1 − (x/a) 2          (7.15)
                            σ y = σ #
                                    1 − (a/x) 2         E
   253   254   255   256   257   258   259   260   261   262   263