OBRABOTKAMETALLOV Vol. 23 No. 3 2021 MATERIAL SCIENCE EQUIPMENT. INSTRUMENTS 4 1 2 Results and Discussion The calculation of the probability of material removing in the presence of vibrations in any area of the contact zone with a known initial state of the surface is calculated by substituting the indicator ( , ) a y a from expression (14) into equation (4) taking into account the parameter , for each of the cases when the initial phase is: 0(2 ) y (15); y (16); 0(2 ) y (17); 0(2 ) (18). For clarity of the calculation procedure, let’s consider a numerical example. Let’s calculate the probability of not removing and the probability of removing the material when grinding holes with a diameter of 150 mm in workpieces made of titanium alloy VT3-1 with a tool AW 60 × 25 × 13 63C F90 M 7 BA 35 m/s (at a wheel speed of 35 m/s, a workpiece speed of 0.25 m/s, longitudinal feed – 33 mm/s, transverse feed – 0.005 mm/stroke). From the calculation of the balance of displacements [20], we determine that for the given processing conditions 6 11.54 10 m f t . Based on the research data [20, 21, 22], we accept: = 0.9 m c K ; 6 = 7.31 10 m z ; 2 = 15.86 grains/mm z n . For the considered conditions 4 =3.397 10 m y L , 628 rad/s , 100 Hz . The calculation is performed according to equations (2), (3), (4) for the level 6 10.38 10 m y at 0.8 2 y L z , = 0.2 f A t . Let’s calculate the parameters y u z V and y y u L V for the cases when the initial phase is equal to 0 y : 4 628( 0.136 10 ) = 0= 0.341, 0.25 –4 628 3.397 10 0 0.853 0.25 Отсюда получим sin 0.3344 , cos 0.942 , sin = 0.753 , cos 0.658 , sin2 = 0.63 , sin2 = 0.991 . After substituting the numerical values of the parameters in (15), we obtain: -6 -6 4 3 3 3 4 5 3 5 0 2 2(11.54 10 10.36 10 ) (3.397 10 ) ( 0.136 10 ) (3.397 10 ) ( 0.136 10 ) 3 0.1 5 0.1 Y 2 2 3 4 6 6 6 0.136 10 +3.397 10 2.308 10 +2 11.54 10 10.38 10 2 6 2 6 2 4 3 2 (2.308 10 ) 0.25( 0.63+0.991) 4 2.308 10 0.25 3.397 10 0.942 ( ) ±0.136 10 0.658 4 628 0.1 628 6 2 2 6 6 3 2 2.308 10 0.25 2 0.25 + 0.1 628 11.54 10 10.38 10 ( 0.3344 + 0.753) 0.1 2 ) 6 8 ( 6 4 2 3 2 15 2 2.308 10 0.25 (3.397 10 ) 0.3344 ( 0.136 10 ) 0.753 = 1.22 10 0.1 628 . Then, according to equation (14), we calculate the value of the indicator, taking into account vibrations ( , ) ( , ) a y a y z : 6 6 15 6 3/2 3 3.14 15.866 10 1 2 7.31 10 (35±0.25)1.22 10 ( , ) = 0.282 8 0.25 (11.54 10 ) a y z .
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