Introduction General part

020. Vertical drilling operation.
Mechanical processing of drilling a hole E in the given detail V surface Ø5H8 mm L = 6 mm long. Detail material Gray cast iron CCH10-35 HB 100-163 parma and its geometric dimensionsL: Spiral parma D = 5 mm, cutting material, hard alloy BK8. Geometric dimensions: 2 ; 2 ; ;
[ 203 page, 44 tab ] for D<16 mm ψ=30
We define the cutting rhythm. ([7] by reference).

1. 
   Determine the depth of cut. T = h = 3.5 mm when removing the deposit amount with one pass.We determine the push value. (25 tab, 277 p)
   

S_o=0.55 mm/round.
We accept So=0.56 mm/round adjusted according to the machine passport.

2. 
   Let's determine the period of stagnation of the cutter.
   

In this case, taking into account that T = 30 ... 60 minutes when working with a single cutter, we assume that T = 45 minutes. ([2],280 p 30 tab )

3. 
   Let’s determine the speed of the main movement at the intersection. (m/min, 265p).
   

;
Here:
From Table 28 (278 b) we record the coefficients and exponents of the formula.
As a cutting tool we use a T15K6 cutter made of hard alloy plate.
C_v=40.4, q=0.45 y_v=0.30, m=0.20
Let's take a look at the correction factors.
([1] 1-t.261p)
n_v=1.7

4. 
   We calculate the frequency of spindle rotations.
   

Adjusting the frequency of revolutions according to the machine passport, we get the actual rotation frequency n = 250 min^-1.
6. The actual speed of the main movement during the cutting process:
7. Determine the torque:
;
;
;
Here: For gray cast iron, the hardness is HB 163, while the cutting material of the drill bit is BK8. [2] (281 page 32-tab.) basically we have:
For torque moments:
=0.01; x=-; q=2.2; y=0.8;
The force acting on the cut
=32.8; q=1.2;x=-; y=0.75.
In that case :
= = 18.3 H/m
= =62147.8 H/m
Shear strength:
= 0.46 kvt;
Main time:
T = = min
Here:
L= =9+2+2=13 mm;

5.