category: Technology

look at the essays like "The calculation and design of gearbox to operating a lathe"

Moscow State Technological University "STANKIN?

CHAIR BASICS design of machines

Course project

on the topic: Calculate and design the gearbox to operatinglathe.

Option 2 / 11

Completed: student gr. BT-6-1 Tulayev PA

Check: Stepanov AA

Moscow 2001

Given:

Tvyh max = 138 H (m nmin = 340 min -1

? = 1,41 n0 = 1000 min -1

type friction clutch EM

type transmission (U = 1) or the clutch on the output shaft of the V-belt

The box set on the stove cast

Lifetime boxes tch = 12 (103 hours

Kinematic calculation

Choice electric

T = 9550 p / p

Calculated output power
Pout =

Rel '=

pobsch = p2op (p2prpobsch = 0, 9952 (0, 982 0 990025 (0, 9604 = 0, 95082

Rel '= (this)

Engine type: Type of performance:
4A132S6U3 M300

pH = 5, 5 kW

n0 = 1000

PP = 965 min-1

engine type selected correctly

Determining the frequency of rotation of the output shaftn1 min = 340 RPM"2 min = n1 (? = 340 (1.41 = 479.4 rpm

definition of common ratios
Uobsch 1, 2 = Upr2 13 (Upr1
Uobsch 1 = (1) = Upr1 (Upr2
Uobsch 2 = (2) = Upr1 (Upr3

choice of gear ratios of individual pairs
Upr max = 4

Partition Uobsch the stairs lead to Uobsch min
Here you can identify the following pairs:

?=

Determination of numbers of teeth of spur gears

?because aw = const

Check rotation frequencies
?
?

%because the kinematics chosen satisfactorily

mm - diameter pulleys at the outputpz = min
30.965> 24 (nII
with TII and pII? bd = 0,3 - calculated transfer

Determination of power in the walls

Rel = 5,5 kW
PI = Pel (? Pr (? Op = 5.5 (0.98 (0.995 = 5.36 kW
RII = PI (? Pr (? Op = 5.36 (0.98 (0.995 = 5.23 kW
RIII = PII (? Op (? Kl.r = 5.23 (0.995 (0.96 = 4.995 kW

Determining the frequency of rotation of shaftsnI = nH (= 965 (= 675.5 min-1nII1 = nI (= 675,5 (= 337,75 min-1nII2 = nI (= 675,5 (= 482,499 min-1nIII1 = nII1 (U = 337,75 min-1nIII2 = nII2 (U = 482,499 min-1


Determination of rotational moments

T = 9550
Tal = 9550 (= 9550 (= 51,103 H (m
TI = 9550 (= 9550 (= 75,7 H (m
TII = 9550 (= 9550 (= 147,8? Tmax = 138 H (m

Engineering Design calculation shafts

? =dbI = 110 (= 32.8 mmdbII = 110 (= 38.8 mmdbIII = 110 (= 35.09 mm

summary table


| ? shaft | Pi | ni | Ti | dbi |
| E | 5,5 | 965 | 51.103 | 38 |
| I | 5,36 | 337,75 | 75,7 | 32,830 |
| II | 5,23 | 482,499 | 147,8 | 38,80 |
| III | 4,995 | 482,499 | 138 | 35,08 |

Calculation spur transmission

because a gear Z3 smallest number of teeth (zmin), then we expectit =

Engineering Design calculation

a) on the contact enduranced1H = Kd (
Kd = 770 (steel)
TI = 75,7 H (m
? bd = 0,3 - ratio of width of the tooth
KH? = 1.07 from Table 1.5
HB> 350
> 6 (less stiff shaft)

Cos? = 1 because spur channel

further on Table 6.5
St40h + heat treatment, hardening in HD
? HP = 900 MPa
? FP = 230 MPa
? HP =? NR '(KHL = 900 (1 = 900MPa
NHO = 8 (107 cycles
NFO = 4 (106 cyclest14 = t24 =

NHE = 60 (tch (nI = 60 (6 (103 (675.5? 24 (107 cycles
KHL = = 1because NHE> NHO, then KHL = 1

dIH = = mmmH = mm

b) on bending endurancemF =
Km = 13,8 (steel, straight-cut)
TI = 75,7 H (m
Z3 = 24
? bd = 0,3
UF3 = Z3 and "X" = 3.92 (see Table)
? Fp =? Fp '(KFL
KFL = 1
KF? = 1.15 from Table 1. 5

For continuous operation
NFE = NHE = 24 (107because NFE> NF0, then KFL = 1
? FP = 230 (1 = 230 MPamF = 13,82,7 mmmH = 2,55 mm mF = 2,7 mm
GOST: 2,0, 2,25, 2,5, 2,75, 3,0, 3,5 ...GOST choose 2,75 mm

checking calculation spur transmission

a) on the contact endurance
? H = ZM (ZH (Z?? HP
ZM = 192 (steel-steel)
ZH = 2,49 (x = 0, t = 0)
Z? =
? = = 1,88-3,2 (() = 1,68
Z? = 0,88dIII =b =? bd (dI = 0,3 (66 = 19,8 mm (taking b = 20)
U = 2
FtI =
KH? = 1 (straight-channel)
KH? = 1,07
KHv =
FHv =? H (q0 (v (b
? H = 0,014 (for spur HB> 350, and without modification)q0 = 47 (for 7 th degree of accuracy)vI =aw =
FHv = 0,014 (47 (2,33 (19,8 (= 213,5 H
KHv = 1 +
? H = 192 (2,49 (0,88 (MPa
730MPa <900MPa

calculation of bending endurance

? F = UFI (R (U? (? FP
UFI = 3,92
Y? = 1 (spur)
Y? = 1 (t = 0)
FtI = 2336 Hb = 19,44 mmm = 2,75 mm
KF? = 1 (spur)
KF? = 1,15
KFv = 1 +
FFv =? F (q0 (vI (b (
? F = 0,016 (direct, without modifying HB> 350)
FFv = 0,016 (47 (2,33 (20 (= 246 H
KFv = 1 + = 1,09
? F = 3,92 (1 (1 (= 205 MPa
205 MPa <230 MPa
SF = = 1,12

Calculation of V-belt transmission
Belt type B
Normal section GOST GOST 1284.1 and 1284.3

characteristics and dimensions (see Table 9.13)B0 = 17 mmBP = 14 mmh = 10,5 mm
A1 = 138 mm2d1min = 125 mmq = 0,18 kg / m
L = 800 ... 6300 mm
T1 = 50 ... 150 Nm

Sheave diameters
mm - diameter pulleys at the outputrounds on the table. 9. 3 to the values of 160 mmdp1 = dp2 = 160 mm

actual speed of output shaftn2 = 482.499 min-1

speed belt

V = 4 m / s


District force
Ft = = 1189 N

Axle base
?mmwhere amin

length belt

L?

L? mm
Accepted standard length of the belt on the table 9.14
L = 1000 mm

final axle distance


Where

? = L -? (Dsr = 497,6

dsr = 160 mm

= 0

mm


The smallest center distance

(required for installation of belt)anaim? a - 0,01 (L? 238,8 mm

largest axle distance

(necessary to compensate for the extraction belt)anaib? a + 0,025 (L? 273,8 mm


Ratio mode


Cp = 1 because lathe (according to Table. 9.9)

angle girth strap on the small pulley


factor angle girth
Ca = 1 (according to Table. 9.15)


The frequency of runs of the belt, with -1

i =

i =

equivalent diameter driving pulley

de = d1 (Ci, where

= 1


=> De = 160 mm

given voltage como

[? F] = 2,5 MPa


Permissible voltage como

[? F] = [? F] 0 (Ca (Cp = 2,5 (1 = 2,5 MPa


The required number of V-belts

Z '=

ratio uneven load distribution along the belts
Sz = 0.95 (according to Table. 9.19)

number of belts

take Z = 3


Ratio regime in one shift

Cp '= 1 (the Table. 9.9)


Working draft ratio

? = 0, 67 (Ca (Cp '= 0,67 (1 (1 = 0,67
Coefficient m =


Cross section belts

A = A1 (Z
A = 138 (3 = 414 mm


The tension from the centrifugal forces

Fts = 10-3 (? (A (V2, where
Density belts? = 1,25 g/cm3
Fts = 10-3 (1,25 (414 (42 = 8,28 N

tension branches at work

F1 = Ft (+ Fts
F2 = Ft (+ Fts
F1 = 1,189 (+8,28 = 1490,13 H
F2 = 1,189 (+8,28 = 301,13 H


Tension branches alone

F0 = 0,5 ((F1 + F2)-x (Fts, wherecoefficient x = 0,2
F0 = 0,5 ((1490,13 +301,13) -0,2 (8,28 = 893,974 H


The forces acting on the transmission shafts at work


Fa = 1774,7 H


The forces acting on the shafts at rest

Fa0 = 2 (F0 (sin
Fa0 = 2 (893,974 (sin 1787,9 H


Sizes profile grooves on the pulleys

(selected according to Table. 9.20)
H = 15
B (b) = 4,2t = 19f = 12,5
? = 34 ? ... 40 ?


Outer diameter pulleys

de1 = de2 = dp1, 2 +2 (bde1, 2 = 168 +2 (4.2 = 176.4 mm

inner diameter pulleys

df1 = df2 = de1, 2 -2 (Hdf1, 2 = 176.4 - 2 (15 = 146,4 mm


The width of the belt


B = Z (t

B = 3 (19 = 57 mm


Width pulley

M = 2 (f + (Z-1) (t
M = 2 (12,5 + (3-1) (19 = 63 mm

Determination of geometric parameters

di =dai = di +2 mdti = di-2, 5mb =? bd (di

d1 = mmda1 = 82,5 +2 (2,75 = 88 mmdt1 = 82,5-2,5 (2,75 = 75,625 mmb1 = 0,3 (82,5 = 24,75 mm

d2 = mmda2 = 115,5 +2 (2,75 = 121 mmdt2 = 115,5-2,5 (2,75 = 108,625 mmb2 = 0,3 (115,5 = 34,65 mm

d3 = mmda3 = 66 +2 (2.75 = 71.5 mmdt3 = 66-2,5 (2,75 = 59,125 mmb3 = 0,3 (66 = 19,8 mm

d4 = mmda4 = 132 +2 (2.75 = 137.5 mmdt4 = 132-2,5 (2,75 = 125,125 mmb4 = 0,3 (132 = 39,6 mm

d5 = mmda5 = 82,5 +2 (2,75 = 88 mmdt5 = 82,5-2,5 (2,75 = 75,625 mmb5 = 0,3 (82,5 = 24,75 mm

d6 = mmda6 = 115,5 +2 (2,75 = 121 mmdt6 = 115,5-2,5 (2,75 = 108,625 mmb6 = 0,3 (115,5 = 34,65 mm

aw = 99 mm (all wheel)

| dt | di | da |

Definition of existing efforts in gearing


Tel = 51,103 H (m
H
H

T1 = TI = 75,7 H (m
H
H

Selection and calculation of the couplings


Electromagnetic clutch with contact studs and the constantnumber of disk type ETM ... 2.



(= 1,3 ... 1,75 coefficient of adhesion
[P] p - specific pressure
[P] p = [P] (Kv
Kv =
Vcp =
JEM =f = 0,25 ... 0,4 (steel Ferodo)-dry
[P] = 0,25 ... 0,3 MPa-dry
T = 75,7 H / mi = 2 (Znar = 2 (3 = 6n = 337,75 rpm
Ar = 53 mm
Tw = 45 mm
JEM =
Vcp =
P =
Kv =
Kv (1
[P] p = 4,17 (0,9 = 3,75

P


More NEWS:
{related-news}