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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 **operating****lathe**.

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

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