Introduction General part

Cutting mode calculation.

005. Horizontal central milling operation.
Milling machine MR72 two surfaces Ø32 va 36mmli the side of the hole surfaces are milled. Cutting tool-torque milling cutter. Installation - special pneumatic mounting. L(Ø32) =90mmcircular surface is being milled.
Casting for processing h =3,7mm Cutting material T15K6.
Side milling diameters D , is calculated depending on the width (V) of the surface to be machined. [11]
D =1,6 В =1,6· 36 =57,6mm.
Standart bo’yicha - standart freza diametri D= 60mm
tishlari soni Z= 4 ta qabul qilamiz.

1. 
   1) Determine the geometric elements (parameters) of the cutter.
   

From reference books [ 2]
= 45....90⁰ taken in borders
We accept =60⁰ (angle in the plan)
_{= 12}⁰ (back angle)
= -5⁰ (front angle)
=+5⁰ (slope angle)
2) determine the depth of cut
the cast is taken in a single worker walk, i.e.t= h =3,7mm
2) We set the cutter to fit one tooth, ( s-2 table) [ 2]
Cutting material - T15K6. Workbench power when N_d =10 kVt
S_z=0,18....0,22 mm/tooth cut.
when S_z=0,2 mm/tooth
3) We determine the stability of the cutter according to the normative document.
The material of the cutting part of the cutter -T15K6 , decay of the tooth along the posterior surface when h_z =1,2mm,
T=180 mm we accept.
4) Determining the cutting speed
a) we determine the cutting speed from the normative document.
The material of the cutting part of the cutter T15K6 , diameter of milling tool D= 60mm,
Z= 4 ta, to t=5mm, to S_z=0,24 mm/tooth

υ_jad =194 m/min.

Taking into account the coefficients of filling speed

Steel 45 mark MPa (=67kGs/mm²) K_m =1,12 and in packaging processing, considering K_p =0,9 ,
υ_{x =} υ_jad K_m K_m =194·1,12 ·0,9=195,5 m/min.
5). We determine the number of rotations of the machine spindle
n= 1000 υ_x /¶d = 1000·195,6 /3,14·80 = 909,7 ayl/min

we adjust the number of revolutions

n_x= 900 ayl/min
6) We determine the true amount of cutting speed

υ_x = ¶d n_x /1000 =3,14·80·900/1000 = 226 m/min
7) We determine the amount of push per minute.
S_m= S_z· Z· n_x=0,2· 4·900 =720 m/min
8) We determine the power expended on the cut
N_jad =6,3 kVt;
N _n = 0,95 (s-6 map) [ 2]
N_kes= N_jad·N _n =6,3·0,95= 6,0 kVt
9) Check the power of the bench drive.
For that
N_kes N_shp
N_shp the strength of the bench spindle N_shp=N_d
On the device N_d =10 kVt, cosidering =0,8
N_shp=N_d =10·0,8 = 8,0 kVt.
Calculation of main time.

t_a = L_iyu / S_m = 60 /720 0.25 =0,3 min

Calculate working walk length.
L_iyu = L+ u+ _m =44 + 12+4= 60 mm
Here:
L – surface length,mm
u = 0,3D=0,3·80=24 :2-= 12mm
_m=1....5 mm cut
010. Horizontal milling machine 6R80
In the given detail, let the surface B be milled in black with a length of 14H8 mm L = 90 mm
In the given detail, the AV surface should be black milled with a length of 14H8 mm L = 90 mm. Horizontal milling machine model 6R80G. The amount of deposit left for processing is h = 2 mm. The surface roughness after machining is Rz = 3.2 μm. Zagotovka material Gray cast iron grade CCH10-35 HB 100-163, hardness 217 HB.
Cutter and its geometric dimensions:
milling cutter material BK6, number of milling teeth z = 16, D=14 mm, milling geometric elements:
We define the cutting rhythm. (According to the reference [7]).
1. We determine the depth of cut. When removing the deposit amount with one pass t=h=2 mm.
2. We determine the push value. (34 t, 283 p)
S_o=0.15 mm/round.
We adjust So = 0.15 mm / month according to the machine passport.
3. Let's determine the period of stagnation of the cutter.
In this case, taking into account that T = 30 ... 180 minutes when working with a single cutter, we assume that T = 180 minutes. ([2], 290 b 40 jad)
4. Let’s determine the speed of the main movement at the intersection. (m/min, 265p).
;
Correction coefficient K
Here:
From Table 39 (287 p) 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=994, q=0.22 ,x=0,17, y_v=0.1,i=0,22, p=0 m=0.33 ([3]288 bet, 39 jad)
Let's take a look at the correction factors.
([3] 1-j.261b) n_v=1.7
V= =13,294 m/min
6. We calculate the frequency of spindle rotations.

Adjusting the frequency of revolutions according to the machine passport, we get the actual frequency of revolutions n = 63min^-1.
7. The actual speed of the main movement during the cutting process:

8. Calculation of the force acting on the shear;
P ;
[3] (291 bet 41-jad) basically we have the following:
Correction coefficient;
For shear strength: K =( ) =1,04 n=0,75
C ; x=01; y=0,75; u=1,1 q=1,3; w=0,2;
P = 1,04= 366,5N
9. We calculate the torque;
M
M 14,80 N m
10. Shear strength:
= kvt;
11. Teeth push speed;
S=zns=16·63·0,15=151,2 mm/min
12.Basic time:
T = min
015. Expansion operation
Let the EOD surfaces be expanded by d = 32 mm l = 38 mm black direction. The amount of deposit left for processing is h = 1.5 mm. The surface roughness after machining is Rz = 40 μm. Zagotovka material Choyan brand 15-32, hardness 160 HB. Cutter and its geometric elements:
Lathe expansion cutter, cutting part material VK6, cutting body material Steel 40, conical cutting surface 25x25 mm l=200 mm
Geometric elements;([4] 187 page, 29 tab )
[188 page, 30 table]; [188 page, 30 tab]; [31 tab]; ,
, , r=1 mm [190 page, 31 tab ]

1. 
   Determine the depth of cut.
   

T = h = 1.5 mm when removing the deposit amount with one pass.
2.Surish qiymatini aniqlaymiz. ([5] 646 page, )
S_o=0.25 mm/round.
We adjust So = 0.25 mm /round according to the machine passport.
3. 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 = 60 minutes. ([5],646 p)
4. Determine the speed of the main movement at the intersection
If t to 2,5 mm, if push S=0,25 mm/ min , for steel, if T15K6 , if ; Vj=218 m/ min
Correction coefficient K=1 ([5]648 page,7 tab)
5. Calculate the frequency of spindle rotations.
Adjusting the frequency of revolutions according to the passport of the machine, we get the actual frequency of revolutions n = 1600 min^-1.
6. The actual speed of the main movement during the cutting process:

7. 
   Power required for cutting:
   

If S=0,25 mm/round, t=1,5mm , Nj=3,4 Kvt ([5] 650 page,7 tab)
Correction coefficient k=0,75
N=3,4 0,75=2,55 Kvt

8. Let's check if the drive is strong enough:

N_shp=N_d·h=10·0.75=7.5 kvt;
N_kes≤N_shp, 2,55≤7.5, that is, it can be processed.
9. We calculate the base time.

Here L=l+y+Δ=95+2+2=99 mm
= =0,2475 min