Calculation of the reducer method. Coursework: Calculation of the gearbox. Calculation without taking into account resistance

Design Assignment 3

1. Selection of the electric motor, kinematic and power calculation of the drive 4

2. Calculation gear wheels gearbox 6

3. Preliminary calculation of the gearbox shafts 10

4. GEARBOX ARRANGEMENT 13

4.1. Constructive dimensions of gears and wheels 13

4.2. Constructive dimensions of the gearbox housing 13

4.3 Gearbox layout 14

5. SELECTION AND CHECK OF BEARING LIFE, SUPPORT REACTIONS 16

5.1. Drive shaft 16

5.2. Output shaft 18

6. RESERVE OF FATIGUE STRENGTH. Revised Shaft Calculation 22

6.1. Drive shaft 22

6.2. Output shaft: 24

7. Calculation of keys 28

8 SELECTING LUBRICANT 28

9 GEARBOX ASSEMBLY 29

REFERENCES 30

Design assignment

Design a single-stage horizontal helical helical gearbox to drive to a belt conveyor.

Kinematic diagram:

1. Electric motor.

2. Electric motor coupling.

3. Gear.

4. Wheel.

5. Drum clutch.

6. Drum of the belt conveyor.

Technical requirements: power on the conveyor drum P b = 8.2 kW, drum rotation frequency n b = 200 rpm.

1. Selection of an electric motor, kinematic and power calculation of the drive

Efficiency of a pair of cylindrical gears η s = 0.96; coefficient taking into account the loss of a pair of rolling bearings, η PC = 0.99; Coupling efficiency η m = 0,96.

Overall drive efficiency

η total =η m 2 ·η PC 3 ·η s = 0.97 2 0.99 3 0.96 = 0.876

Power on the drum shaft R b = 8.2 kW, n b= 200 rpm. Required motor power:

R dv =
=
= 9.36 kW

N dv = n b(2 ... 5) =
= 400 ... 1000 rpm

Choosing an electric motor based on the required power R dv= 9.36 kW, three-phase squirrel-cage motor, series 4A, closed, blown, with a synchronous speed of 750 rpm 4A160M6U3, with parameters R dv= 11.0 kW and slip 2.5% (GOST 19523-81). Rated engine speed:

n dv= rpm.

Ratio i= u= n nom / n b = 731/200=3,65

Determine the rotational speed and angular velocities on all drive shafts:

n dv = n nom = 731 rpm

n 1 = n dv = 731 rpm

rpm

n b = n 2 = 200.30 rpm

where is the frequency of rotation of the electric motor;

- rated speed of the electric motor;

- frequency of rotation of the high-speed shaft;

- the frequency of rotation of the low-speed shaft;

i= u - gear ratio of the reducer;

- the angular speed of the electric motor;

- angular speed of the high-speed shaft;

- the angular speed of the low-speed shaft;

- the angular speed of the driving drum.

Determine the power and torque on all drive shafts:

R dv = P demand = 9.36 kW

R 1 = P dv ·η m = 9.36 0.97 = 9.07 kW

R 2 = P 1 ·η PC 2 ·η s = 9.07 0.99 2 0.96 = 8.53 kW

R b = P 2 · η m ·η PC = 8.53 0.99 0.97 = 8.19 kW

where
- electric motor power;

- power on the pinion shaft;

- power on the wheel shaft;

- power on the drum shaft.

Determine the torque of the electric motor and the torques on all drive shafts:

where - the torque of the electric motor;

- torque of the high-speed shaft;

- the torque of the low-speed shaft;

- torque of the drive drum.

2. Calculation of gears of the reducer

For the gear and wheel, we select materials with average mechanical characteristics:

For gear, steel 45, heat treatment - improvement, hardness HB 230;

For the wheel - steel 45, heat treatment - improvement, hardness HB 200.

We calculate the permissible contact stresses using the formula:

,

where σ H lim b- the limit of contact endurance at the base number of cycles;

TO HL- coefficient of durability;

- safety factor.

For carbon steels with tooth surface hardness less than HB 350 and heat treatment (improvement)

σ H lim b = 2НВ + 70;

TO HL accept equal 1, because the projected service life is more than 5 years; safety factor = 1.1.

For helical gears, the calculated permissible contact stress is determined by the formula:

for gear
= MPa

for wheel =
MPa.

Then the calculated permissible contact stress

Condition
done.

The center distance from the conditions of contact endurance of the active surfaces of the teeth is found by the formula:

,

where
- the hardness of the surfaces of the teeth. For a symmetrical arrangement of wheels relative to the supports and with a material hardness of ≤350HB, we take in the range (1 - 1.15). Let's take = 1.15;

ψ ba = 0.25 ÷ 0.63 is the ratio of the width of the crown. We accept ψ ba = 0.4;

K a = 43 - for helical and chevron gears;

u - ratio. and = 3,65;

.

Accept center distance
, i.e. round to the nearest whole number.

The normal modulus of engagement is taken according to the following recommendation:

m n =
=
mm;

we accept in accordance with GOST 9563-60 m n= 2 mm.

Let us preliminarily accept the angle of inclination of the teeth β = 10 о and calculate the number of teeth of the gear and wheel:

Z1 =

We accept z 1 = 34, then the number of teeth of the wheel z 2 = z 1 · u= 34 3.65 = 124.1. We accept z 2 = 124.

We clarify the value of the angle of inclination of the teeth:

The main dimensions of the gear and wheel:

dividing diameters:

Examination:
mm;

tooth tip diameters:

d a 1 = d 1 +2 m n= 68.86 + 22 = 72.86 mm;

d a 2 = d 2 +2 m n= 251.14 + 22 = 255.14 mm;

diameters of the cavities of the teeth: d f 1 = d 1 - 2 m n= 68.86-2 * 2 = 64.86 mm;

d f 2 = d 2 - 2 = 251.14-2 * 2 = 247.14 mm;

determine the width of the wheel : b2=

determine the width of the gear: b 1 = b 2 + 5mm = 64 + 5 = 69mm.

Determine the ratio of the width of the gear by diameter:

The peripheral speed of the wheels and the degree of transmission accuracy:

At this speed, for helical wheels, we take the 8th degree of accuracy, where the load factor is:

TO Hβ we take it equal to 1.04.

since the hardness of the material is less than 350HB.

Thus, K H = 1.04 1.09 1.0 = 1.134.

We check the contact voltages according to the formula:

We calculate the overload:

Overload is within normal limits.

Forces acting in engagement:

district:

;

radial:

where
= 20 0 - the angle of engagement in the normal section;

= 9.07 0 is the angle of inclination of the teeth.

We check the teeth for endurance by bending stresses using the formula:

.

,

where
= 1.1 - coefficient taking into account the uneven distribution of the load along the length of the tooth (load concentration coefficient);

= 1.1 - coefficient taking into account the dynamic action of the load (dynamic coefficient);

Factor that takes into account the shape of the tooth and depends on the equivalent number of teeth

Allowable stress according to the formula

.

For improved steel 45 with hardness HB≤350 σ 0 F lim b= 1.8 HB.

For gear σ 0 F lim b= 1.8 230 = 415 MPa; for wheel σ 0 F lim b= 1.8 200 = 360 MPa.

= ΄˝ - safety factor, where ΄ = 1.75, ˝ = 1 (for forgings and stampings). Therefore,. = 1.75.

Allowable voltages:

for gear
MPa;

for wheel
MPa.

Finding an attitude
:

for gear
;

for wheel
.

Further calculation should be carried out for the teeth of the wheel, for which the found ratio is less.

Determine the coefficients Y β and K Fα:

where TO Fα- coefficient taking into account the uneven distribution of the load between the teeth;

=1,5 - end overlap ratio;

n = 8 is the degree of accuracy of the gears.

We check the strength of the wheel tooth using the formula:

;

The strength condition is met.

3. Preliminary calculation of the gearbox shafts

The diameters of the shafts are determined by the formula:

.

For the drive shaft [τ to] = 25 MPa; for the slave [τ to] = 20 MPa.

Drive shaft:

For a 4A engine, 160M6U3 = 48 mm. Shaft diameter d in 1 =48

Let's take the diameter of the shaft under the bearings d n1 = 40 mm

Coupling diameter d m = 0.8 =
= 38.4 mm. We accept d m = 35 mm.

The free end of the shaft can be determined by the approximate formula:

,

where d NS – diameter of the shaft for the bearing.

Under the bearings we take:

Then l=

The schematic design of the drive shaft is shown in Fig. 3.1.

Rice. 3.1. Drive shaft design

Driven shaft.

Output shaft end diameter:

, we take the closest value from the standard series

We take under the bearings

Under the cogwheel

The schematic design of the driven (low-speed) shaft is shown in Figure 3.2.

Rice. 3.2. Output shaft design

The diameters of the remaining sections of the shafts are assigned based on design considerations when assembling the gearbox.

4. ARRANGEMENT OF THE GEARBOX

4.1. Constructive dimensions of gears and wheels

We carry out the gear in one piece with the shaft. Its dimensions:

width

diameter

tooth tip diameter

cavity diameter
.

Forged wheel:

width

diameter

tooth tip diameter

cavity diameter

hub diameter

hub length,

accept

Rim Thickness:

accept

Disc thickness:

4.2. Constructive dimensions of the gearbox housing

Thickness of the walls of the case and lid:

We accept

We accept
.

Thickness of flanges of body and cover belts:

upper body belt and cover belt:

lower belt of the body:

We accept
.

Bolt diameter:

fundamental; we accept bolts with M16 thread;

fastening the cover to the housing at the bearings

; we accept bolts with M12 thread;

connecting the cover to the body; we accept bolts with M8 thread.

4.3 Gearbox layout

The first stage serves to approximately determine the position of the gears relative to the supports for the subsequent determination of the support reactions and the selection of bearings.

The layout drawing is carried out in one projection - a section along the axes of the shafts with the gearbox cover removed; scale 1: 1.

Gearbox housing dimensions:

we take the gap between the end of the gear and the inner wall of the body (if there is a hub, we take the gap from the end of the hub); we take A 1 = 10 mm; in the presence of a hub, the clearance is taken from the end of the hub;

we take the gap from the circumference of the tops of the teeth of the wheel to the inner wall of the body
;

we take the distance between the outer ring of the drive shaft bearing and the inner wall of the housing; if the diameter of the circumference of the tops of the gear teeth is greater than the outer diameter of the bearing, then the distance must be taken from the gear.

Preliminarily, we outline deep groove ball bearings single row of the middle series; the dimensions of the bearings are selected according to the shaft diameter at the bearing seat
and
.(Table 1).

Table 1:

Dimensions of the intended bearings

	Bearing designation				Lifting capacity, kN
		dimensions, mm
Fast
Slow-moving

We are solving the issue of lubricating the bearings. We accept grease for bearings. To prevent the lubricant from leaking out into the housing and the washing out of the grease by liquid oil from the engagement zone, we install grease-retaining rings.

A sketchy layout is shown in Fig. 4.1.

5. SELECTION AND CHECK OF BEARING LIFE, SUPPORT REACTIONS

5.1. Drive shaft

From the previous calculations we have:

We define support reactions.

The design diagram of the shaft and the diagrams of bending moments are shown in Fig. 5.1

In the YOZ plane:

Examination:

in the XOZ plane:

Examination:

in the YOZ plane:

section 1:
;

section 2: M
=0

Section 3: M

in the XOZ plane:

section 1:
;

=

section2:

section3:

We select the bearing according to the most loaded support. We outline the radial ball bearings 208: d=40 mm;D=80mm; V=18mm; WITH= 32.0 kN; WITH O = 17.8kN.

where R B= 2267.3 N

- temperature coefficient.

Attitude
; this value corresponds
.

Attitude
; X = 0.56 andY=2,15

Estimated durability according to the formula:

where
- rotational speed of the drive shaft.

5.2. Driven shaft

The driven shaft carries the same loads as the driving one:

The design diagram of the shaft and the diagrams of bending moments are shown in Fig. 5.2

We define support reactions.

In the YOZ plane:

Examination:

In the XOZ plane:

Examination:

Total reactions in supports A and B:

We determine the moments by areas:

in the YOZ plane:

section 1: at x = 0,
;

at x= l 1 , ;

section 2: at x= l 1 , ;

at x =l 1 + l 2 ,

section 3 :;

in the XOZ plane:

section 1: at x = 0,;

at x= l 1 , ;

section 2: at x =l 1 + l 2 ,

section 3: at x= l 1 + l 2 + l 3 ,

We build diagrams of bending moments.

We select the bearing according to the most loaded support and determine their durability. We outline the radial ball bearings 211: d=55 mm;D=100mm; V=21mm; WITH= 43.6 kN; WITH O = 25.0 kN.

where R A= 4290.4 N

1 (the inner ring rotates);

Safety factor for belt conveyor drives;

Temperature coefficient.

Attitude
; this value corresponds to e = 0.20.

Attitude
, then X = 1, Y = 0. That's why

Estimated durability, mln.

Estimated durability, h

where
- the frequency of rotation of the driven shaft.

6. RESERVE OF FATIGUE STRENGTH. Refined shaft calculation

Let us assume that normal voltages bends change in a symmetric cycle, and tangents from torsion change in a pulsating one.

The refined calculation of shafts consists in determining the safety factors s for dangerous shaft sections and comparing them with the required values ​​[s]. The strength is observed at
.

6.1 Drive shaft

Section 1: at x = 0,;

at x =l 3 , ;

Section 2: at x =l 3 , ;

at x =l 3 + l 2 , ;

Section 3: at x =l 3 + l 2 , ;

at x =l 3 + l 2 + l 1 , .

Torque:

We define dangerous sections. To do this, we schematically depict the shaft (Fig. 8.1)

Rice. 8.1 Schematic representation of the drive shaft

Two sections are dangerous: under the left bearing and under the gear. They are dangerous because complex stress state (bending with torsion), significant bending moment.

Stress concentrators:

1) the bearing is seated in a transitional fit (pressing is less than 20 MPa);

2) fillet (or groove).

Determine the safety factor for fatigue strength.

With a workpiece diameter up to 90mm
average value of tensile strength for 45 steel with heat treatment - improvement
.

Symmetrical bending cycle endurance limit:

Endurance limit at symmetric shear stress cycle:

Section A-A. The stress concentration is due to the bearing fit with guaranteed interference:

Because press-on pressure is less than 20 MPa, then we reduce the value of this ratio by 10%.

for the steels mentioned above, we take
and

Bending moment from diagrams:

Axial moment of resistance:

Amplitude of normal voltages:

Average voltage:

Polar moment of resistance:

Amplitude and average stress of the cycle of shear stresses according to the formula:

Safety factor for normal stresses according to the formula:

Safety factor for shear stresses according to the formula:

The resulting coefficient is greater than the permissible norms (1.5 ÷ 5). Therefore, the diameter of the shaft must be reduced, which in this case should not be done, because Such a large safety factor is due to the fact that the diameter of the shaft was increased during the design to connect it with a standard coupling to the electric motor shaft.

6.2. Driven shaft:

Determine the total bending moments. The values ​​of the bending moments for the sections are taken from the diagrams.

Section 1: at x = 0,;

at x =l 1 , ;

Section 2: at x =l 1 , ;

at x =l 1 + l 2 , ;

Section 3: at x =l 1 + l 2 , ; .

Amplitude and average stress of the cycle of shear stresses:

Safety factor for normal stresses:

Safety factor for shear stresses:

The resulting safety factor for the section according to the formula:

Because the resulting safety factor under the bearing is less than 3.5, then there is no need to reduce the shaft diameter.

7. Calculation of keys

The material of the keys is steel 45 normalized.

Collapse stress and strength condition are determined by the formula:

.

Maximum shear stress with a steel hub [ σ cm ] = 100120 MPa, with cast iron [ σ

We set the viscosity of the oil. At contact voltages
= 400.91 MPa and speed
the recommended oil viscosity should be approximately equal to
We accept industrial oil I-30A (according to GOST 20799-75).

9. ASSEMBLY OF THE GEARBOX

Before assembly, the inner cavity of the gearbox housing is thoroughly cleaned and coated with oil-resistant paint.

The assembly is carried out in accordance with the assembly drawing of the gearbox, starting from the shaft assemblies:

on the drive shaft lubricating rings and ball bearings, preheated in oil up to 80-100 0 С;

a key is laid in the driven shaft
and press on the gear wheel until it stops in the collar of the shaft; then put on a spacer sleeve, oil retaining rings and install ball bearings, preheated in oil.

The assembly of shafts is placed in the base of the gearbox housing and the housing cover is put on, having previously coated the surface of the joint between the cover and the housing with alcohol varnish. For centering, install the cover on the body using two tapered pins; tighten the bolts securing the cover to the body.

After that, grease is placed in the bearing chambers of the driven shaft, bearing caps are installed with a set of metal gaskets for adjustment.

Before setting the through covers, rubber reinforced cuffs are placed in the grooves. Check the absence of jamming of the bearings by turning the shafts and fix the covers with bolts.

Then screw in the oil drain plug with a gasket and a rod pointer.

Pour oil into the body and close the inspection hole with a cover with a gasket made of technical cardboard; fix the cover with bolts.

The assembled gearbox is run in and tested at the stand according to the program established by the technical conditions. The calculation of the calculations is summarized in Table 2: Table 2 Geometric parameters low-speed stage of cylindrical reducer Options...

Design and verification payment reducer

Coursework >> Industry, Manufacturing

There is an electric motor selection, design and test payment reducer and his component parts... V ... Output: ΔU = 1% gearbox [ΔU] = 4%), kinematic payment performed satisfactorily. 1.4 Calculation of frequencies, powers ...

1. Choosing an electric motor

Gearbox kinematic diagram:

1. Engine;

2. Reducer;

3. Drive shaft;

4. Safety clutch;

5. The coupling is elastic.

Z 1 - worm

Z 2 - worm wheel

Determination of drive power:

First of all, we choose an electric motor, for this we determine the power and speed.

The power consumption (W) of the drive (output power) is determined by the formula:

transmission electric motor drive

Where Ft is the circumferential force on the drum of the conveyor belt or the sprocket of the apron conveyor (N);

V is the speed of the chain or belt (m / s).

Electric motor power:

Where z total is the overall efficiency of the drive.

z total = z m? z h.p z m z pp;

where h.p is the efficiency of the worm gear;

z m - coupling efficiency;

z p3? efficiency of bearings of the 3rd shaft

s total = 0.98 0.8 0.98 0.99 = 0.76

I determine the power of the electric motor:

2. Determination of the rotational speed of the drive shaft

drum diameter, mm.

According to the table (24.8), we select the "air132m8" electric motor

with rotation frequency

with power

torque t max / t = 2,

3. Determination of the total gear ratio and its breakdown by stages

Choose from the standard range

We accept

Check: Suitable

4. Determination of power, speed and torque for each shaft

5. Determination of allowable stresses

Determine the sliding speed:

(From paragraph 2.2 calculation of gears) we take V s> = 2 ... 5 m / s II tinless bronzes and brass, taken at a speed

Total running time:

The total number of cycles of voltage alternation:

Worm. Steel 18 KhGT is case-hardened and hardened to HRC (56 ... 63). Coils are ground and polished. ZK profile.

Worm wheel. The dimensions of the worm pair depend on the value of the permissible stress [y] H for the material of the worm wheel.

Allowable stresses for calculating the strength of working surfaces:

Material of the 2nd group. Bronze Br АЖ 9-4. Casting into the ground

y in = 400 (MPa); y t = 200 (MPa);

Because both materials are suitable for the manufacture of a gear rim, then we choose a cheaper one, namely Br AZ 9-4.

I accept a worm with the number of calls Z 1 = 1, and a worm wheel with the number of teeth Z 2 = 38.

I determine the initial permissible stresses for calculating the teeth of the worm wheel for the strength of the working surfaces, the bending endurance limit of the material of the teeth and the safety factor:

y F o = 0.44? y t + 0.14? y b = 0.44 200 + 0.14 400 = 144 (MPa);

S F = 1.75; K FE = 0.1;

N FE = K FE N? = 0.1 34200000 = 3420000

I determine the maximum allowable voltage:

[y] F max = 0.8? y t = 0.8 200 = 160 (MPa).

6. Load factors

I determine the approximate value of the load factor:

k I = k v I k in I;

k in I = 0.5 (k in o +1) = 0.5 (1.1 + 1) = 1.05;

k I = 1 1.05 = 1.05.

7. Determination of the design parameters of the worm gear

The preliminary value of the center distance:

With a constant load factor K I = 1.0 K hg = 1;

T not = K ng ChT 2;

K I = 0.5 (K 0 I +1) = 0.5 (1.05 + 1) = 1.025;

Tinless bronzes (material II)

At К he at the solution of loading I is equal to 0.8

Accept a" w = 160 (mm).

I define the axial module:

I accept the module m= 6.3 (mm).

Worm Diameter Coefficient:

Accept q = 12,5.

Worm displacement coefficient:

Determine the angles of ascent of the turn of the worm.

Pitch pitch angle:

8. Checking calculation of the worm gear for strength

Load concentration factor:

where And - the coefficient of deformation of the worm;

X is a coefficient that takes into account the influence of the transmission operating mode on the running-in of the worm wheel teeth and the worm turns.

for the 5th loading mode.

Load factor:

k = k v k in = 1 1.007 = 1.007.

Sliding speed in engagement:

Allowable voltage:

Design voltage:

200.08 (MPa)< 223,6 (МПа).

The calculated stress on the working surfaces of the teeth does not exceed the allowable one, therefore, the previously established parameters can be taken as final.

Efficiency:

I clarify the value of the power on the worm shaft:

I determine the forces in the engagement of the worm pair.

Circumferential force on the wheel and axial force on the worm:

Circumferential force on the worm and axial force on the wheel:

Radial force:

F r = F t2 tgb = 6584 tg20 = 2396 (H).

Bending stress in the teeth of the worm wheel:

where Y F = 1.45 is a coefficient that takes into account the shape of the teeth of the worm wheels.

18.85 (MPa)< 71,75 (МПа).

Transmission test for short-term peak load.

Peak moment on the worm wheel shaft:

Peak contact stress on the working surfaces of the teeth:

316.13 (MPa)< 400 (МПа).

Peak bending stress of the worm gear teeth:

Checking the gearbox for heating.

Heating temperature installed on the metal frame of the gearbox with free cooling:

where t o - ambient air temperature (20 о С);

kt - heat transfer coefficient, kt = 10;

A - area of ​​the cooling surface of the gearbox housing (m 2);

A = 20 a 1.7 = 20 0.16 1.7 = 0.88 (m 2).

56.6 (o C)< 90 (о С) = [t] раб

Since the heating temperature of the gearbox during free cooling does not exceed the permissible value, artificial cooling is not required for the gearbox.

9. Definition geometric dimensions worm gear

Pitch diameter:

d 1 = m q = 6.3 12.5 = 78.75 (mm).

Initial diameter:

d w1 = m (q + 2x) = 6.3 (12.5 + 2 * 0.15) = 80.64 (mm).

Diameter of the tops of the turns:

d a1 = d 1 + 2m = 78.75 + 2 6.3 = 91.35 = 91 (mm).

The diameter of the hollows of the turns:

d f1 = d 1 -2h * f m = 78.75-2 1.2 6.3 = 63.63 (mm).

Length of the threaded part of the worm:

c = (11 + 0.06 z 2) m + 3 m = (11 + 0.06 38) 6.3 + 3 6.3 = 102.56 (mm).

We accept b = 120 (mm).

Worm wheel.

Pitch and initial diameter:

d 2 = d w2 = z 2 m = 38 6.3 = 239.4 (mm).

Tooth tip diameter:

d a2 = d 2 +2 (1 + x) m = 239.4 + 2 (1 + 0.15) 6.3 = 253.89 = 254 (mm).

Tooth cavity diameter:

d f2 = d 2 - (h * f + x) 2m = 239.4 - (1.2 + 0.15) 26.3 = 222.39 (mm).

Crown width

in 2 ? 0.75 d a1 = 0.75 91 = 68.25 (mm).

We accept 2 = 65 (mm).

10. Determination of shaft diameters

1) The diameter of the high-speed shaft is accepted

We accept d = 28 mm

Shaft chamfer size.

Bearing seating diameter:

We accept

We accept

2) Diameter of low-speed shaft:

We accept d = 45 mm

For the found shaft diameter, select the values:

Approximate bead height,

Maximum radius of the bearing chamfer,

Shaft chamfer size.

Determine the diameter of the bearing seating surface:

We accept

Bearing shoulder diameter:

Accept:.

10. Selection and check of rolling bearings by dynamic load rating

1. For the high-speed shaft of the gearbox, we will choose single-row angular contact ball bearings of the middle series 36307.

For him we have:

Inner ring diameter

Outer ring diameter

Bearing width,

The bearing is affected by:

Axial force,

Radial force.

Rotation frequency:.

Required work resource :.

Safety factor

Temperature coefficient

Rotation ratio

Let's check the condition:

2. For the low-speed gearbox shaft, we will choose single-row angular contact ball bearings of the light series.

For him we have:

Inner ring diameter

Outer ring diameter

Bearing width,

Dynamic load capacity,

Static load capacity,

Limiting speed with grease.

The bearing is affected by:

Axial force,

Radial force.

Rotation frequency:.

Required work resource :.

Safety factor

Temperature coefficient

Rotation ratio

Axial loading ratio :.

Let's check the condition:

Determine the value of the coefficient of radial dynamic load x = 0.45 and the coefficient of axial dynamic load y = 1.07.

Determine the equivalent radial dynamic load:

Let's calculate the resource of the accepted bearing:

Which meets the requirements.

12. Calculation of the drive shaft (the most loaded) shaft for fatigue strength and endurance

Active loads:

Radial force

Torque -

A moment on a drum

Let us determine the reactions of the supports in the vertical plane.

Let's check:,

Consequently, vertical reactions are found correctly.

Let us determine the reactions of the supports in the horizontal plane.

we get that.

Let's check the correctness of finding the horizontal reactions:, - true.

The moments in the dangerous section will be equal:

The calculation is carried out in the form of checking the safety factor, the value of which can be taken. In this case, the condition must be fulfilled that, where is the calculated safety factor, and are the safety factors for normal and shear stresses, which we will define below.

Find the resulting bending moment as.

Determine the mechanical characteristics of the shaft material (Steel 45): - tensile strength (tensile strength); and - the limits of endurance of smooth specimens with a symmetric cycle of bending and torsion; - coefficient of material sensitivity to stress cycle asymmetry.

Let's define the ratio of the following quantities:

where and are the effective coefficients of stress concentration, is the coefficient of influence of the absolute dimensions of the cross section. Let us find the value of the coefficient of influence of roughness and the coefficient of influence of surface hardening.

Let us calculate the values ​​of the stress concentration factors and for a given section of the shaft:

Determine the limits of shaft endurance in the section under consideration:

Let's calculate the axial and polar moments of resistance of the shaft section:

where is the calculated shaft diameter.

We calculate the bending and shear stress in the dangerous section using the formulas:

Determine the safety factor for normal stresses:

To find the safety factor for shear stresses, we define the following values. The coefficient of influence of the asymmetry of the stress cycle for a given section. Average cycle voltage. Let's calculate the safety factor

Let's find the calculated value of the safety factor and compare it with the admissible one: - the condition is met.

13. Calculation of keyway connections

The calculation of keyed connections consists in checking the condition of the strength of the key material for crushing.

1. Key on the low-speed shaft for the wheel.

We accept a key 16x10x50

Strength condition:

1. Key on low-speed shaft for coupling.

Shaft torque, - shaft diameter, - key width, - key height, - shaft groove depth, - hub groove depth, - allowable collapse stress, - yield strength.

Determine the working length of the key:

We accept a key 12x8x45

Strength condition:

14. Choice of couplings

To transfer torque from the motor shaft to the high-speed shaft and to prevent the shaft from skewing, we select a coupling.

For the drive of the belt conveyor, the most suitable is an elastic coupling with a toroidal shell in accordance with GOST 20884-82.

The clutch is selected depending on the torque on the low-speed shaft of the gearbox.

Toric-shell couplings have high torsional, radial and angular flexibility. The half-couplings are installed on both cylindrical and tapered shaft ends.

Values ​​of displacements of each type admissible for this type of couplings (provided that displacements of other types are close to zero): axial mm, radial mm, angular. The loads acting on the shafts can be determined from the graphs from the literature.

15. Lubrication of the worm gear and bearings

A crankcase system is used to lubricate the transmission.

Determine the peripheral speed of the tops of the teeth of the wheel:

For a low-speed stage, here is the frequency of rotation of the worm wheel, is the diameter of the circumference of the tops of the worm wheel

Let us calculate the maximum permissible level of immersion of the gear of the low-speed gear stage in the oil bath:, here is the diameter of the circles of the tops of the teeth of the high-speed gear stage.

Let us determine the required volume of oil by the formula:, where is the height of the oil filling area, and are the length and width of the oil bath, respectively.

Let's choose the brand of oil I-T-S-320 (GOST 20799-88).

And - industrial,

T - heavily loaded nodes,

C - oil with antioxidants, anti-corrosion and anti-wear additives.

The bearings are lubricated with the same oil by splashing. When assembling the gearbox, the bearings must be pre-oiled.

Bibliography

1. P.F. Dunaev, O. P. Lelikov, "Design of units and machine parts", Moscow, "High School", 1985.

2. D.N. Reshetov, "Machine parts", Moscow, "Mechanical engineering", 1989.

3. R.I. Gzhirov, "A quick reference of the designer", "Mechanical engineering", Leningrad, 1983.

4. Atlas of structures "Machine parts", Moscow, "Mechanical engineering", 1980.

5. L. Ya. Perel, A.A. Filatov, reference book "Rolling bearings", Moscow, "Mechanical engineering", 1992.

6. A.V. Boulanger, N.V. Palochkina, L. D. Chasovnikov, guidelines for the calculation of gears of reducers and gearboxes for the course "Machine parts", part 1, Moscow, MSTU im. N.E. Bauman, 1980.

7. V.N. Ivanov, V.S. Barinov, "Selection and calculations of rolling bearings", guidelines for course design, Moscow, MSTU im. N.E. Bauman, 1981.

8.E.A. Vitushkin, V.I. Strelov. Calculation of gear shafts. MSTU them. N.E. Bauman, 2005.

9. Atlas of "structures of units and machine parts", Moscow, publishing house MSTU im. N.E. Bauman, 2007.

The purchase of a motor gearbox is an investment in technical and technological business processes, which should not only be justified, but also payback. And the payback largely depends on selection of a geared motor for specific purposes. It is carried out on the basis of a professional calculation of power, dimension, productive efficiency, the required load level for specific purposes of use.

To avoid mistakes that can lead to early wear and tear of equipment and costly financial losses, calculation of the geared motor must be produced by qualified personnel. If necessary, it and other studies to select a gearbox can be carried out by experts of the PTC "Privod" company.

Selection by main characteristics

Long service life while maintaining the specified level of performance of the equipment with which it operates is a key benefit in the right choice drive. Our long-term practice shows that when defining requirements, it is worth proceeding from the following parameters:

• at least 7 years of maintenance-free work for the worm gear;
• from 10-15 years for a cylindrical drive.

In the course of determining the data for placing an order for manufacture of gear motor the key characteristics are:

• power of the connected electric motor,
• the speed of rotation of the moving elements of the system,
• type of motor power supply,
• operating conditions of the gearbox - operating mode and loading.

At calculating the power of the electric motor for the geared motor the performance of the equipment with which it will work is taken as a basis. The performance of a geared motor largely depends on the output torque and the speed of its operation. The speed, like the efficiency, can change with voltage fluctuations in the motor power supply system.

The speed of the geared motor is a dependent variable influenced by two characteristics:

• ratio;
• the frequency of rotational movements of the motor.

Our catalog contains gearboxes with different speed parameters. Models available with one or more speed modes... The second option provides for the presence of a regulation system speed parameters and is used in cases when, during the operation of the gearbox, it is necessary to periodically change the speed modes.

Motor power supply - is carried out through the supply of a constant or alternating current... Motor gearboxes direct current designed for connection to a network with 1 or 3 phases (under voltage 220 and 380V, respectively). AC drives operate at 3, 9, 12, 24 or 27V.

The professional, depending on the operating conditions, requires a determination of the nature and frequency / intensity of future use. Depending on the nature of the loaded activity for which the gearbox is designed, it can be a device:

• for work in bumpless mode, with moderate or strong impacts;
• with a smooth starting system to reduce destructive loads when starting and stopping the drive;
• for continuous operation with frequent starts (by the number of starts per hour).

According to the operating mode, the geared motor can be designed for long-term operation of the engine without overheating in especially heavy, heavy, medium, light duty.

Selection according to the type of gearbox for the drive

A professional calculation in order to select a gearbox always begins with a study of the drive circuit (kinematic). It is she who underlies the compliance of the selected equipment with the conditions of future operation. According to this diagram, you can choose the class of the geared motor. The options are as follows.

• :
  • single stage transmission, input shaft at right angles to output shaft (crossed position of input shaft and output shaft);
  • a two-stage mechanism with the input shaft parallel or perpendicular to the output shaft (axes can be vertical / horizontal).
• :
  • with a parallel position of the input shaft and output shaft and horizontal placement of the axes (the output shaft with the input organ are in the same plane);
  • with the placement of the axes of the input shaft and output in the same plane, but coaxially (located at any angle).
• Conical-cylindrical. In it, the axis of the input shaft intersects with the axis of the output shaft at an angle of 90 degrees.

When choosing a geared motor, the position of the output shaft is of key importance. With an integrated approach to the selection of a device, the following should be considered:

• Cylindrical and conical motor reducer, having similar weight and dimensions to a worm drive, demonstrates a higher efficiency.
• The load transmitted by a spur gearbox is 1.5–2 times higher than that of a worm gear.
• The use of bevel and spur gears is only possible when placed horizontally.

Classification by the number of stages and the type of transmission

Reducer type	Number of steps	Transfer type	Axis arrangement
Cylindrical	1	One or more cylindrical	Parallel
	2		Parallel / coaxial
	3
	4		Parallel
Conical	1	Conical	Intersecting
Conical-cylindrical	2	Conical Cylindrical (one or more)	Intersecting / Interbreeding
	3
	4
Worm	1	Worm (one or two)	Interbreeding
	2		Parallel
Cylinder-worm or worm-cylindrical	2	Cylindrical (one or two) Worm (one)	Interbreeding
	3
Planetary	1	Two central gear wheels and satellites (for each step)	Coaxial
	2
	3
Cylindrical planetary	2	Cylindrical (one or more) Planetary (one or more)	Parallel / coaxial
	3
	4
Bevel planetary	2	Conical (one) Planetary (one or more)	Intersecting
	3
	4
Planetary worm	2	Worm (one) Planetary (one or more)	Interbreeding
	3
	4
Wave	1	Wave (one)	Coaxial

Ratio

Determination of the gear ratio is performed according to a formula of the form:

U = n in / n out

• n in - the revolutions of the input shaft (characteristic of the electric motor) per minute;
• n out - the required number of revolutions of the output shaft per minute.

The resulting quotient is rounded to the gear ratio from the standard range for specific types of geared motors. The key condition for a successful choice of an electric motor is the limitation on the speed of the input shaft. For all types of drive mechanisms, it should not exceed 1.5 thousand revolutions per minute. The specific frequency criterion is specified in technical characteristics engine.

Gear ratio range for gearboxes

Capacities

With the rotational movements of the working bodies of the mechanisms, resistance arises, which leads to friction - abrasion of the nodes. With the right choice of the gearbox in terms of power, it is able to overcome this resistance. Because this moment is of great importance when you need buy gear motor with long-term goals.

The power itself - P - is considered as a quotient of the force and speed of the gearbox. The formula looks like this:

• where:
  M - moment of force;
• N - revolutions per minute.

To select the required gear motor, it is necessary to compare the data on the power at the input and output - P1 and P2, respectively. Calculation of the power of the geared motor the output is calculated as follows:

• where:
  P is the power of the reducer;
  Sf - service factor, also known as service factor.

The output of the reducer (P1> P2) must be lower than the input. The rate of this inequality is explained by the inevitable loss of productivity when engaging as a result of friction between parts.

When calculating capacities, it is imperative to use accurate data: due to different efficiency indicators, the probability of selection error when using approximate data is close to 80%.

Efficiency calculation

The efficiency of the geared motor is the quotient of the division of the output and input power. Calculated as a percentage, the formula is:

ñ [%] = (P2 / P1) * 100

When determining efficiency, one should rely on the following points:

• the value of the efficiency directly depends on the gear ratio: the higher it is, the higher the efficiency;
• during the operation of the gearbox, its efficiency may decrease - it is affected by both the nature or operating conditions and the quality of the lubricant used, adherence to the schedule scheduled repairs, timely service, etc.

Reliability indicators

The table below shows the resource standards for the main parts of the geared motor during long-term operation of the device with constant activity.

Resource

Buy gear motor

PTC "Privod" is a manufacturer of gearboxes and geared motors with different characteristics and efficiency, which is not indifferent to the indicators of the payback of its equipment. We are constantly working not only to improve the quality of our products, but also to create the most comfortable conditions for its purchase for you.

To minimize selection errors, our customers are offered an intelligent one. You don't need any special skills or knowledge to use this service. The tool works online and will help you determine the optimal type of equipment. We will offer the best geared motor price of any type and full support of its delivery.

The presence of the drive kinematic diagram will simplify the choice of the gearbox type. Gearboxes are structurally divided into the following types:

Gear ratio [I]

The gear ratio of the gearbox is calculated by the formula:

I = N1 / N2

where
N1 - shaft rotation speed (number of rpm) at the input;
N2 - shaft rotation speed (rpm) at the output.

The calculated value is rounded up to the value specified in the technical data for a specific type of gearbox.

Table 2. Range gear ratios for different types of gearboxes

IMPORTANT!
The rotation speed of the electric motor shaft and, accordingly, the input shaft of the gearbox cannot exceed 1500 rpm. The rule applies to all types of gearboxes, except for cylindrical coaxial gearboxes with a rotation speed of up to 3000 rpm. This technical parameter manufacturers indicate in the summary characteristics of electric motors.

Gearbox torque

Output torque- torque on the output shaft. The rated power, safety factor [S], estimated operating time (10 thousand hours), gearbox efficiency are taken into account.

Rated torque- maximum torque ensuring safe transmission. Its value is calculated taking into account the safety factor - 1 and the duration of operation - 10 thousand hours.

Maximum torque (M2max]- the limiting torque that the gearbox can withstand under constant or varying loads, operation with frequent starts / stops. This value can be interpreted as an instantaneous peak load in the operating mode of the equipment.

Required torque- torque that meets the customer's criteria. Its value is less than or equal to the rated torque.

Calculated torque- the value required to select the gearbox. The calculated value is calculated using the following formula:

Mc2 = Mr2 x Sf ≤ Mn2

where
Mr2 is the required torque;
Sf - service factor (operating factor);
Mn2 is the rated torque.

Service factor (service factor)

Service factor (Sf) is calculated experimentally. The calculation takes into account the type of load, the daily operating time, the number of starts / stops per hour of operation of the geared motor. The service factor can be determined using the data in Table 3.

Table 3. Parameters for calculating the service factor

Load type	Number of starts / stops, hour	Average duration of operation, days
		<2	2-8	9-16h	17-24
Soft start, static operation, medium mass acceleration	<10	0,75	1	1,25	1,5
	10-50	1	1,25	1,5	1,75
	80-100	1,25	1,5	1,75	2
	100-200	1,5	1,75	2	2,2
Moderate Start Load, Variable Mode, Medium Mass Acceleration	<10	1	1,25	1,5	1,75
	10-50	1,25	1,5	1,75	2
	80-100	1,5	1,75	2	2,2
	100-200	1,75	2	2,2	2,5
Heavy Duty, Variable Duty, Large Mass Acceleration	<10	1,25	1,5	1,75	2
	10-50	1,5	1,75	2	2,2
	80-100	1,75	2	2,2	2,5
	100-200	2	2,2	2,5	3

Drive power

Correctly calculated drive power helps to overcome the mechanical frictional resistance that occurs during straight and rotary movements.

The elementary formula for calculating power [P] is the calculation of the ratio of force to speed.

For rotary movements, power is calculated as the ratio of torque to rpm:

P = (MxN) / 9550

where
M - torque;
N is the number of revolutions / min.

The output power is calculated using the formula:

P2 = P x Sf

where
P - power;
Sf is the service factor (operating factor).

IMPORTANT!
The input power value must always be higher than the output power value, which is justified by the meshing losses:

P1> P2

Calculations cannot be made using an approximate input power, as efficiency can vary significantly.

Coefficient of performance (COP)

We will consider the calculation of efficiency using the example of a worm gear. It will be equal to the ratio of mechanical output power and input power:

ñ [%] = (P2 / P1) x 100

where
P2 - output power;
P1 is the input power.

IMPORTANT!
In worm gear P2< P1 всегда, так как в результате трения между червячным колесом и червяком, в уплотнениях и подшипниках часть передаваемой мощности расходуется.

The higher the gear ratio, the lower the efficiency.

The efficiency is influenced by the service life and the quality of the lubricants used for preventive maintenance of the gearmotor.

Table 4. Efficiency of a single-stage worm gearbox

Ratio	Efficiency at a w, mm
	40	50	63	80	100	125	160	200	250
8,0	0,88	0,89	0,90	0,91	0,92	0,93	0,94	0,95	0,96
10,0	0,87	0,88	0,89	0,90	0,91	0,92	0,93	0,94	0,95
12,5	0,86	0,87	0,88	0,89	0,90	0,91	0,92	0,93	0,94
16,0	0,82	0,84	0,86	0,88	0,89	0,90	0,91	0,92	0,93
20,0	0,78	0,81	0,84	0,86	0,87	0,88	0,89	0,90	0,91
25,0	0,74	0,77	0,80	0,83	0,84	0,85	0,86	0,87	0,89
31,5	0,70	0,73	0,76	0,78	0,81	0,82	0,83	0,84	0,86
40,0	0,65	0,69	0,73	0,75	0,77	0,78	0,80	0,81	0,83
50,0	0,60	0,65	0,69	0,72	0,74	0,75	0,76	0,78	0,80

Table 5. Efficiency of the wave reducer

Table 6. Efficiency of gear reducers

Explosion-proof versions of geared motors

Gear motors of this group are classified according to the type of explosion-proof design:

• "E" - units with an increased degree of protection. They can be operated in any mode of operation, including emergency situations. Enhanced protection prevents the possibility of ignition of industrial mixtures and gases.
• "D" - flameproof enclosure. The housing of the units is protected against deformation in the event of an explosion of the gearmotor itself. This is achieved due to its design features and increased tightness. Equipment with explosion protection class "D" can be used in extremely high temperatures and with any groups of explosive mixtures.
• "I" is an intrinsically safe circuit. This type of explosion protection provides support for a non-explosive current in the electrical network, taking into account the specific conditions of industrial application.

Reliability indicators

Reliability figures for geared motors are shown in table 7. All values ​​are given for continuous operation at constant rated load. The geared motor must provide 90% of the resource indicated in the table even in the mode of short-term overloads. They occur when the equipment is started and the rated torque is doubled, at least.

Table 7. Resource of shafts, bearings and gears of gearboxes

For the calculation and purchase of geared motors of various types, please contact our specialists. you can familiarize yourself with the catalog of worm, cylindrical, planetary and wave gear motors offered by Techprivod.

Romanov Sergey Anatolyevich,
head of mechanics department
company Tekhprivod.

Other helpful materials:

There are 3 main types of geared motors - planetary, worm and helical geared motors. To increase the torque and further reduce the speed at the output of the geared motor, there are various combinations of the above types of geared motors. We suggest you use calculators for an approximate calculation of the power of the geared motor of the mechanisms for lifting the load and mechanisms for moving the load.

For lifting mechanisms.

1. Determine the required speed at the output of the geared motor based on the known lifting speed

V = π * 2R * n, where

R- radius of the lifting drum, m

V-lifting speed, m * min

n- revolutions at the output of the geared motor, rpm

2.determine the angular speed of rotation of the shaft of the geared motor

3.determine the required effort to lift the load

m is the mass of the cargo,

g- acceleration due to gravity (9.8m * min)

t- coefficient of friction (somewhere 0.4)

4. Determine the torque

5.calculate the power of the electric motor

Based on the calculation, we select the required gear motor from the technical specifications on our website.

For mechanisms for moving cargo

Everything is the same, except for the formula for calculating the effort

a - acceleration of the load (m * min)

T is the time during which the cargo travels along, for example, a conveyor

For load lifting mechanisms, it is better to use MCH, MRCH gear motors, since they exclude the possibility of the output shaft spinning when an effort is applied to it, which eliminates the need to install a shoe brake on the mechanism.

For the mechanisms of mixing mixtures or drilling, we recommend planetary gear motors 3Mp, 4MP as they experience uniform radial load.