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Allowances calculation.
In the given detail we calculate the number of main deposits and intermediate boundary dimensions on the sequence of mechanical drilling of the surface T, drilling length Ø5H8 mm L = 6 mm.
Half-finished product. Rz = 200 mkm T = 300 mkm. ([1] page 63, table 4,3)
In drilling Rz=40 mkm ,T=60 mkm. ([1] page 65, table 4,5)
In countersinking Rz=50 mkm ,T=50 mkm.
When calculating the inside surface of the allowance.
2z ([1] page 62,table 4,2)
The total value of spatial deviations for a given blank is determined by the following formula:
We determine the convexity of surface B.
ρkor=∆k·L
∆k=0,8 L=6 mm
ρkor=∆k·L=0,8·6 =4.8 mkm
ρsm can be obtained by making the thickness of the casting wall equal to the tolerance:
ρsm =430 mkm.
= =430.1 mkm.
The residual spatial deviation after drilling is:
ρ2=ρ1·0,05=430·0,05=21.5 mkm.
Let's find out if there are any installation errors.
Installation error:
=0.07 mm,
=0.43 mm. [2] (page 441, table 2)
0.25 mm.
= 125 mkm. [1] (page 79, table 4.12)
In that case:
= =279.5 mkm
Based on the values entered in the table, we calculate the values of the intermediate, minimum transitions from the transitions using the following formula:
;
Amount of minimum allowance.
Drilling
2 =2(200+300+ =2027 mkm.
Countersinking
2 =2(40+60+ =265.9 mkm.
Let's find the calculated size.
D=9H8+0.07=9.07 mm.
Let's find the calculated size.
D=9.07-0.265=8.805 mm.
D=8.805-2.027=6.778 mm.
Calculated dimensions:
D=8.805+0.07=8.875 mm .
D=6.778+0.43=7.208 mm.
Limit amounts of deposits:
Total allowance:
Checking.
δzag-δ1 =0.43-0.07 =0.36 mkm.
The calculation is correct.
In the given detail, we calculate the amount of deposits and intermediate boundary dimensions based on the sequence of machining by black clean milling with a width of B = 85 mm L = 95 mm, keeping the surface A in size h = 95h8.
Half-finished product Rz=200 mkm T=300 mkm. ([1] page 63, table 4,3)
Primary Rz=50 mkm ,T=50 mkm. ([1] page 64, table 4,5)
Qo’yimning yuzalarini xisoblashda :
z ([1] page62, table 4,2).
The total value of spatial deviations for a given blank is determined by the following formula:
We determine the coroblenia of surface B.
ρkor=∆k·L
∆k=0,8, L=95 mm
ρkor=∆k·L=0,8·95 =76 mkm
ρsm can be obtained by equal to the thickness of the casting wall dopusk:
ρsm =430 mkm.
= =430,96 mkm.
The residual spatial deviation after drilling is:
ρ2=ρ1·0,05=430,96·0,05=21.54 mkm.
Let's find out if there are any installation errors.
Installation error:
=0.063 mm,
=0.630 mm. [2] (441-bet, 2- jad)
0.346 mm.
= 130mkm. [1] (page 79, table 4.12)
In that case:
= =370 mkm
In clear installation
Based on the values entered in the table, we calculate the values of the intermediate, minimum transitions from the transitions using the following formula:
Minimum deposit amount on the route.
z =200+300+448+370=1318 mkm
We’ll find the calculated size.
H=95H8+0.063=95,063 mm.
We’ll find the calculated sizes.
H=95,063+1,318=96.381 mm
Calculated dimensions:
H=95,063-0.063=95,00 mm .
H=96,381-0.63=95,751 mm.
Boundary quantities of allowances.
Checking.
δzag-δ1 =0.63-0.063 =0.567 mkm.
The calculation is correct.
In the given detail, we calculate the amount of deposits and intermediate boundary dimensions based on the sequence of machining of surface B with a black clean milling machine L = 65 f 10 mm long. Write the parameters of the deposit:
mkm; mkm ( [3] page 44, table 2)
Let's write the minimum: in primary cutting
Z=2000 mkm ([3]136 bet, 30 jad)
We will calculate the given dimensions;
H =65 =65,062 mm
H=65,062+2,00=67,062 mm
Calculated dimensions given:
H =65,062-0,062=65,00 mm
H=67,062-0,25=66,812 mm
We will calculate the quotes:
Zmin=66,812-65.00=1,812 mm
Maximum deposits:
Zmin=67,062-65.062=2 mm
We check the accounts:
Zmax-Zmin=2-1,812=0,188 mm
0,25-0,062=0,188 mm
So the calculations are correct.
Table 3.
Surfaces
|
Surface dimensions
|
Marking
|
Account deposits. Mkm
|
Allowances. Mkm
|
A
|
Ø28mm
|
2Zmin
|
2000
|
=62
=250
|
B
|
Ø20mm
|
2Zmin
|
400
7600
|
=43
=430
|
C, I, L
|
2mm×45°
|
2Zmin
|
2000
|
=62
=250
|
D
|
Ø14mm
|
2Zmin
|
2000
|
=62
=250
|
E
|
36mm
|
Zmin
|
2000
|
=62
=250
|
F
|
M36-7H
|
2Zmin
|
2000
|
=33
=330
|
G, O
|
3mm×45°
|
Zmin
|
400
7600
|
=52
=520
|
H
|
Ø32mm
|
2Zmin
|
2000
|
=33
=330
|
J
|
41mm
|
Zmin
|
|
=62
=250
|
K
|
Ø32mm
|
2Zmin
|
2000
|
=33
=330
|
M
|
Ø18mm
|
2Zmin
|
2000
|
=33
=330
|
N
|
M22-7H
|
2Zmin
|
2000
|
=33
=330
|
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