Probabilistic model of surface layer removal when grinding brittle non-metallic materials

OBRABOTKAMETALLOV Vol. 23 No. 2 2021 TECHNOLOGY After substituting the values of the gamma functions, we get:                   2 3 5 0 1 3 2 3 2 ( ) (1 )( ) 2 8 ( , ) 5 15 3 8 c z k u z f y y y u u k V V n P t y z z a y z z L L L V H                    4 9 7 5 3 3/2 2 2 3/2 3 2 ( ) ( ) 4 6 4 8 5 20 3 16 9 7 c z k u z f y y y u u f y y k V V n t y z z z z z L L L V H t L L . (21) The calculation of the indicator 2 ( , ) a y z , characterizing the change in the area of the depressions formed due to the brittle cleavage process in any area of the contact zone with the known initial state of the surface is calculated similarly to the indicator 1 ( , ) a y z . To calculate the indicator 2 ( , ) a y z , it is necessary to take into account that the course of the brittle shearing process is accompanied by an increase in the width of the single risk z b to the value x b (Fig. 4). For approximation x b , a power-law dependence was used:               2 x m x bx f x e z b C t y u r D , (22) where  x r – is the increment in material removal in the process of brittle chipping of brittle non-metallic material; x m – is the exponent in the equation, that simulates the shearing grain pro fi le as a paraboloid of revolution. The depth distribution density of shear grains can be calculated using the formula:      1 ( ) x x x u f u u H , (23) where  x – is the parameter of the distribution density function of shear grains. The dependence for calculating the indicator 2 ( , ) a y z , included in the expression for calculating the probability of material removal due to volume brittle fracture, similar to the solution given above (18), is written as:                      ( ) 2 1 2 0 ( ) ( , ) y m t z y z c b k u z f e L u u k C V V n z a y z t y u u dudz D V H                      ( ) 2 0 1 0 ( ) y m t z y z c b k u z f e L u u k C V V n P z t y u u dudz D V H                               ( ) 2 0 1 0 ( ) x x y m t z y r z c b k u z f x e f L u u k C V V n P z u t y u r u dudz D t V H . (24) After integrating expression (24) over u we get:                         2 0 2 ( ) ( ) ( 1)(1 ) ( , ) ( 1) m z c b k u z f e L u u k C V V n m P z a y z t y dz D m V H à à à                             2 ( ) ( ) ( 1) ( 1) x m z c b k u z x f x e L x u u f k C V V n m z t y r dz D m V H t à à à . (25)