Example 1

Determine the gear ratio of the gear train (Fig. 19), the number of revolutions of the driven shaft and the overall coefficient of performance (efficiency), if the number of teeth of the wheels are equal: z 1 =30, z 2 =20, z 3 =45, z 4 =30, z 5 =20, z 6 =120, z 7 =25, z 8 =15 ; number of revolutions of the input shaft n 1 =1600 rpm.

Solution

The mechanism consists of four stages: two cylindrical z 1 - z 2 , z 3 - z 4 with external gear, cylindrical z 5 - z 6 with internal gear and conical z 7 - z 8 .

The total gear ratio of a multi-stage transmission is equal to the product of the gear ratios of each stage that form this gear mechanism. For this case

.

The sign (-) indicates that the direction of rotation of the wheels in these pairs is opposite. The direction of rotation of the wheels in this case can also be determined by placing arrows on the diagram (Fig. 19).

The number of revolutions of the driven shaft is determined through the gear ratio
rpm

The overall efficiency of the gear mechanism is equal to

where the numerical values ​​are taken according to the condition of the problem T1.

Example 2

Here
,
,
- gear ratios of the converted mechanism (carrier H stopped and the fixed wheel is spinning z 3 ). The resulting gear ratio with the “+” sign indicates the coincidence of the directions of rotation of the driving and driven shafts.

Example 3

Solution

As in example 2, this mechanism refers to a single-stage planetary gear and the gear ratio from the carrier H to the wheel z 1 is determined by the relation

Example 4

Solution

A complex gear train consists of two stages: the first stage is a simple cylindrical pair with external gearing z 1 -z 2, the second stage is a planetary gear N-z 5 , transmitting rotational motion from the carrier H to the wheel z 5 via satellite z 4 . The direction of rotation of the output shaft is determined by an algebraic sign.

1. For a two-stage transmission, the total gear ratio is found through the gear ratios of each stage, i.e.

.

Received gear ratio
, which indicates an increase in the frequency of rotation of the output shaft, and the “+” sign indicates that the directions of rotation of the shafts are the same.

2. Determine the angular velocity of the output link and its angular acceleration

rad/s,

rad/s 2 .

3. Since the rotation of the wheels is accelerated (we accept uniformly accelerated), then the time during which the angular velocities will double, we determine from the dependence

,

where And - angular velocities, respectively, at the beginning and at the end of the considered period of time
. From here

from.

4. Determine the overall transmission efficiency

Task T2

The output link of the mechanism shown in the diagrams (Fig. 23–32) performs a reciprocating (or reciprocating) movement and is loaded on the working stroke with a constant force F c (or moment T from) useful resistance. At idle, with the reverse direction of movement of the output link, there is no useful resistance, but harmful ones continue to act. Taking into account the effect of friction in kinematic pairs, in terms of efficiency mechanism to be determined.

1) driving moment T d , constant in magnitude, which must be applied to the input link in steady motion with a cycle consisting of working and idle strokes;

2) the work of the friction forces on the working and idle strokes, considering that the harmful resistance is constant on each of the strokes, but on the working stroke it is three times greater than at idle;

3) change in the kinetic energy of the mechanism during the working stroke and during the idling;

4) the power required from the drive when the input link rotates at an average speed and average (for a whole revolution) power of useful resistance and friction forces.

The solution of this problem is based on the equation of motion of the mechanism, which establishes the relationship between the change in kinetic energy and the work of forces (the law of kinetic energy). The work of forces and moments is determined, respectively, by the linear or angular displacements of the links on which they act. In this regard, it is required to determine the positions of the mechanism at the extreme positions of the output link. Link movements, linear and angular, can be determined from a drawing made to scale or calculated analytically. The dimensions of the links, according to their designations on the mechanism diagram, and other necessary values ​​\u200b\u200bare given in the tables of numerical data, where is the efficiency factor, and in option 9 m- rack and pinion module, z - the number of teeth of the wheel.

Table 17

Value	Penultimate cipher digit
OA, mm
OS, mm
Sun, mm
AB, mm
T from , Nm
, rad/s

Table 18

Value	Penultimate cipher digit
oa, mm
AB, mm
F c , H
, rad/s

Table 19

Value	Penultimate cipher digit
OA, mm
OV, mm
T from , Nm
, rad/s

Table 20

	Value	Penultimate cipher digit
	oa, mm
	OV, mm
	BC=BD, mm
	F c , H
	, rad/s

Table 21

Value	Penultimate cipher digit
R, mm
oa, mm
F c , H
, rad/s

Table 22

Value	Penultimate cipher digit
OA, mm
OV, mm
BD, mm
F c , H
, rad/s

Table 23

Value	Penultimate cipher digit
OA, mm
e, mm
F c , H
, rad/s

Table 24

Value	Penultimate cipher digit
R, mm
oa, mm
r, mm
F c , H
, rad/s

Table 25

Value	Penultimate cipher digit
oa, mm
AB, mm
m, mm
T from, Nm
, rad/s

Table 26

Value	Penultimate cipher digit
oa, mm
OV, mm
F c , H
, rad/s

The sequence of the task. First, it is necessary to build a mechanism in extreme positions, and in given directions of the angular velocity of the input link
and constant strength F from (or moment T from) useful resistance to set working and idling strokes.

When graphically determining the linear and angular displacements of the links, it is necessary to remove from the drawing:

1) for the input link, its rotation angles on the working stroke and at idle X;

2) for the output link during its reciprocating motion, linear displacement, i.e. move s, or during its reciprocating rotational movement, the swing angle
.

In order to determine the zones of working and idle strokes for the input link, it is necessary to take into account the connection of movement with the shown direction of action of useful resistance, which should prevent the movement of the output link during the working stroke.

In options 5 and 8, the links in the upper pair are positively locked, preventing the links from moving away from each other: in option 8, the roller of radius r rolls in the circular groove of the input link, covered by the outer and inner profiles of the groove, in option 5 the round eccentric is covered by the frame of the output link.

INTRODUCTION

Worm gear refers to gears with intersecting shafts.

The main advantages of worm gears: the possibility of obtaining large gear ratios in one pair, smooth engagement, the possibility of self-braking. Disadvantages: relatively low efficiency, increased wear and a tendency to seize, the need to use expensive anti-friction materials for wheels.

Worm gears are more expensive and more complicated than gears, so they are used, as a rule, when it is necessary to transfer motion between intersecting shafts, and also where a large gear ratio is needed.

The criterion for the performance of worm gears is the surface strength of the teeth, which ensures their wear resistance and the absence of chipping and seizing, as well as bending strength. Under the action of short-term overloads in the worm gear, the teeth of the worm wheel are checked for bending according to the maximum load.

For the body of the worm, a verification calculation for stiffness is carried out, and a thermal calculation is also carried out.

The design is carried out in two stages: design - from the conditions of contact endurance, the main dimensions of the transmission are determined and verification - with known parameters of the transmission under the conditions of its operation, contact and bending stresses are determined and compared with those allowed by the endurance of the material.

The forces loading the bearings are determined and the bearings are selected according to their load capacity.

KINEMATIC AND FORCE CALCULATION

Motor selection

To select an electric motor, its required power and speed are determined.

According to the initial design data, the required power to perform the technological process can be found from the formula:

P out \u003d F t V, (2.1)

where P out - power on the output shaft of the drive, W;

F t - traction force, N;

V is the speed of movement of the working body, m/s;

P out \u003d 1.5 kW.

Determination of the overall efficiency drive

Then, in accordance with the kinematic power transmission chain, the total efficiency. of the entire drive is calculated by the formula:

s total = s 1 s 2 s 3 s 4 (2.2)

h total = 0.80.950.980.99 = 0.74.

Thus, based on the overall efficiency. it became clear that during the operation of the drive, only 74% of the power from the engine would go to the winch drum.

Let's determine the required engine power for normal operation of the winch:

We accept a 2.2 kW motor.

Calculation of the rotational speed of the motor shaft

Since at this stage the gear ratios of the drive gears are still unknown and the motor shaft speed is not known, it becomes possible to calculate the desired speed of the motor shaft.

For this, the following calculations were carried out.

Determination of the speed of the output shaft of the drive

According to the initial data, the angular velocity of the output shaft is calculated by the formula:

where u - angular velocity, s -1;

D b - drum diameter, m;

v is the speed of movement of the working body, m/s.

Let's find the rotation frequency, knowing the angular velocity by the formula:

rpm (2.5)

Determining the desired drive ratio

From the analysis of the kinematic diagram of the electric winch drive, it can be seen that its total gear ratio (u total) is formed due to the gear ratio of the worm gear reducer.

We accept u chp = 50. The relationship between the frequencies of rotation of the motor shaft n dv and the output shaft n z is determined by the relationship:

n dv = n z u total, (2.6)

then the desired speed of the motor shaft will be:

n engine = 38.250 = 1910 rpm.

According to the current range of motors, the one closest to the desired speed is a motor with a synchronous speed of 1500 rpm. In view of the foregoing, we finally accept the brand engine: 90L4 / 1395. AIR series, which has the following characteristics:

R dv \u003d 2.2 kW;

n motor = 1500 rpm.

Kinematic calculations

Total gear ratio:

u total \u003d n dv / \u003d 1500 / 38.2 \u003d 39.3.

Let us determine all the kinematic characteristics of the designed drive, which will be needed in the future for a detailed study of the transmission. Determination of frequency and rotation speeds. It is easy to calculate the rotational speeds of all shafts, starting from the selected rotational speed of the electric motor shaft, taking into account the fact that the rotational speed of each subsequent shaft is determined through the rotational speed of the previous one according to the formula (2.7), taking into account the gear ratio:

where n (i+1) - speed i+1 shaft, rpm;

u i -(i+1) - gear ratio between i and i+1 shafts.

Moments on the gearbox shafts:

T 1 \u003d 9.5510 3 (P / n e) \u003d 9.5510 3 (2.2 / 1500) \u003d 14.0 Nm

T 2 \u003d T 1 u \u003d 14.039.3 \u003d 550 Nm.

Worm gear is one of the classes of mechanical gearboxes. Gearboxes are classified according to the type of mechanical transmission. The screw that underlies the worm gear looks like a worm, hence the name.

Gearmotor- this is a unit consisting of a gearbox and an electric motor, which are in one unit. Worm gear motorcreated in order to work as an electromechanical motor in various general purpose machines. It is noteworthy that this type of equipment works perfectly both under constant and variable loads.

In a worm gearbox, the increase in torque and decrease in the angular velocity of the output shaft occurs due to the conversion of energy contained in the high angular velocity and low torque on the input shaft.

Errors in the calculation and selection of the gearbox can lead to its premature failure and, as a result, in the best case to financial loss.

Therefore, the work on the calculation and selection of the gearbox must be entrusted to experienced design specialists who will take into account all factors from the location of the gearbox in space and operating conditions to its heating temperature during operation. Having confirmed this with appropriate calculations, the specialist will ensure the selection of the optimal gearbox for your specific drive.

Practice shows that a properly selected gearbox provides a service life of at least 7 years for worm gearboxes and 10-15 years for cylindrical gearboxes.

The choice of any gearbox is carried out in three stages:

1. Gearbox type selection

2. Selection of the overall size (size) of the reducer and its characteristics.

3. Checking calculations

1. Gearbox type selection

1.1 Initial data:

The kinematic diagram of the drive, indicating all the mechanisms connected to the gearbox, their spatial arrangement relative to each other, indicating the mounting points and methods of mounting the gearbox.

1.2 Determining the location of the axes of the gearbox shafts in space.

Helical gearboxes:

The axis of the input and output shafts of the gearbox are parallel to each other and lie in only one horizontal plane - a horizontal spur gearbox.

The axis of the input and output shaft of the gearbox are parallel to each other and lie in only one vertical plane - a vertical spur gearbox.

The axis of the input and output shaft of the gearbox can be in any spatial position, while these axes lie on the same straight line (coincide) - a coaxial cylindrical or planetary gearbox.

Bevel-helical gearboxes:

The axis of the input and output shaft of the gearbox are perpendicular to each other and lie only in one horizontal plane.

Worm gears:

The axis of the input and output shafts of the gearbox can be in any spatial position, while they cross at an angle of 90 degrees to each other and do not lie in the same plane - a single-stage worm gearbox.

The axis of the input and output shaft of the gearbox can be in any spatial position, while they are parallel to each other and do not lie in the same plane, or they cross at an angle of 90 degrees to each other and do not lie in the same plane - a two-stage gearbox.

1.3 Determination of the mounting method, mounting position and gearbox assembly option.

The method of fastening the gearbox and the mounting position (mounting on the foundation or on the driven shaft of the drive mechanism) are determined according to the technical characteristics given in the catalog for each gearbox individually.

The assembly option is determined according to the schemes given in the catalog. Schemes of "Assembly options" are given in the "Designation of gearboxes" section.

1.4 In addition, the following factors can be taken into account when choosing a gearbox type

1) Noise level

• the lowest - for worm gears
• the highest - for cylindrical and bevel gears

2) Efficiency

• the highest - for planetary and single-stage spur gearboxes
• the lowest - in worm, especially two-stage

Worm gears are preferably used in intermittent operation

3) Material consumption for the same values ​​of torque on a low-speed shaft

• the lowest - for planetary single-stage

4) Dimensions with the same gear ratios and torques:

• the largest axial - in coaxial and planetary
• the largest in the direction perpendicular to the axes - for cylindrical
• the smallest radial - to planetary.

5) Relative cost rub/(Nm) for the same center distances:

• the highest - in conical
• the lowest - in planetary

2. Selection of the overall size (size) of the reducer and its characteristics

2.1. Initial data

Drive kinematic diagram containing the following data:

• type of drive machine (engine);
• the required torque on the output shaft T required, Nxm, or the power of the propulsion system P required, kW;
• frequency of rotation of the input shaft of the gearbox n in, rpm;
• frequency of rotation of the output shaft of the gearbox n out, rpm;
• the nature of the load (uniform or uneven, reversible or irreversible, the presence and magnitude of overloads, the presence of shocks, shocks, vibrations);
• the required duration of operation of the gearbox in hours;
• average daily work in hours;
• the number of starts per hour;
• duration of inclusions with load, PV%;
• environmental conditions (temperature, heat removal conditions);
• duration of inclusions under load;
• radial cantilever load applied in the middle of the landing part of the ends of the output shaft F out and the input shaft F in;

2.2. When choosing the size of the gearbox, the following parameters are calculated:

1) Gear ratio

U= n in / n out (1)

The most economical is the operation of the gearbox at an input speed of less than 1500 rpm, and for the purpose of longer trouble-free operation of the gearbox, it is recommended to use an input shaft speed of less than 900 rpm.

The gear ratio is rounded up to the nearest number according to table 1.

The table selects the types of gearboxes that satisfy the given gear ratio.

2) Calculated torque on the gearbox output shaft

T calc \u003d T required x K dir, (2)

T required - the required torque on the output shaft, Nxm (initial data, or formula 3)

K dir - operating mode coefficient

With a known power of the propulsion system:

T required \u003d (P required x U x 9550 x efficiency) / n in, (3)

P required - power of the propulsion system, kW

n in - the frequency of rotation of the input shaft of the gearbox (provided that the shaft of the propulsion system directly transmits rotation to the input shaft of the gearbox without additional gear), rpm

U - gear ratio of the gearbox, formula 1

Efficiency - efficiency of the gearbox

The operating mode coefficient is defined as the product of the coefficients:

For gear reducers:

K dir \u003d K 1 x K 2 x K 3 x K PV x K roar (4)

For worm gears:

K dir \u003d K 1 x K 2 x K 3 x K PV x K rev x K h (5)

K 1 - coefficient of the type and characteristics of the propulsion system, table 2

K 2 - coefficient of duration of work table 3

K 3 - coefficient of the number of starts table 4

K PV - coefficient of duration of inclusions table 5

K rev - coefficient of reversibility, with non-reversible operation K rev = 1.0 with reverse operation K rev = 0.75

K h - coefficient taking into account the location of the worm pair in space. When the worm is located under the wheel, K h \u003d 1.0, when located above the wheel, K h \u003d 1.2. When the worm is located on the side of the wheel, K h \u003d 1.1.

3) Calculated radial cantilever load on the gearbox output shaft

F out. calculated = F out x K dir, (6)

F out - radial cantilever load applied in the middle of the landing part of the ends of the output shaft (initial data), N

K dir - operating mode coefficient (formula 4.5)

3. The parameters of the selected gearbox must meet the following conditions:

1) T nom > T calc, (7)

T nom - rated torque on the output shaft of the gearbox, given in this catalog in the technical specifications for each gearbox, Nxm

T calc - estimated torque on the output shaft of the gearbox (formula 2), Nxm

2) F nom > F out calc (8)

F nom - rated cantilever load in the middle of the landing part of the ends of the gearbox output shaft, given in the technical specifications for each gearbox, N.

F out.calc - calculated radial cantilever load on the output shaft of the gearbox (formula 6), N.

3) R inlet calc< Р терм х К т, (9)

R in.calc - the estimated power of the electric motor (formula 10), kW

P term - thermal power, the value of which is given in the technical characteristics of the gearbox, kW

K t - temperature coefficient, the values ​​\u200b\u200bof which are given in table 6

The rated power of the electric motor is determined by:

R in.calc \u003d (T out x n out) / (9550 x efficiency), (10)

T out - estimated torque on the output shaft of the gearbox (formula 2), Nxm

n out - the speed of the output shaft of the gearbox, rpm

Efficiency - the efficiency of the gearbox,

A) For spur gearboxes:

• single-stage - 0.99
• two-stage - 0.98
• three-stage - 0.97
• four-stage - 0.95

B) For bevel gears:

• single-stage - 0.98
• two-stage - 0.97

C) For bevel-helical gearboxes - as the product of the values ​​​​of the bevel and cylindrical parts of the gearbox.

D) For worm gearboxes, the efficiency is given in the technical specifications for each gearbox for each gear ratio.

To buy a worm gearbox, find out the cost of the gearbox, choose the right components and help with questions that arise during operation, the managers of our company will help you.

Table 1

table 2

Leading machine	Generators, elevators, centrifugal compressors, evenly loaded conveyors, mixers of liquid substances, centrifugal pumps, gear, screw, boom mechanisms, blowers, fans, filtering devices.	Water treatment facilities, unevenly loaded conveyors, winches, cable drums, running, turning, lifting mechanisms of cranes, concrete mixers, furnaces, transmission shafts, cutters, crushers, mills, equipment for the oil industry.	Punch presses, vibrators, sawmills, screens, single cylinder compressors.	Equipment for the production of rubber products and plastics, mixing machines and equipment for shaped steel.
electric motor, steam turbine
4, 6 cylinder internal combustion engines, hydraulic and pneumatic engines
1, 2, 3 cylinder internal combustion engines

Table 3

Table 4

Table 5

Table 6

cooling	Ambient temperature, C o	Duration of inclusion, PV%.
Reducer without outsider cooling.
Reducer with water cooling spiral.

Algorithm #1

Calculation of closed gear

Cylindrical gear

A l g o r i t m

calculation closed gear spur and helical

cylindrical gear

The terms of reference must contain the following information:

Power on the gear shaft .......... .P 1, kW;

Gear speed ............................... n 1 , rpm;

Wheel speed ............................... n 2 , rpm;

(other parameters can be set, determined by

previous ones);

transmission reversibility;

Transmission service life .................................. t d, years;

Annual usage rate.... K G;

Daily usage rate... K from;

- loading histogram:

Paragraph 1. Preparation of design parameters.

1.1. Preliminary determination of the gear ratio

Coordinate with standard values ​​(Table 1.1). Select nearest standard value U.

Actual output speed

RPM (2)

Deviation from the value of the terms of reference

(3)

1.2. Torque on the gear shaft

1.3. Transmission time

t = t g (years)×365(days)×24(hours)× TO g× TO s, hour. (five)

Point 2. Material selection . Determination of allowable stresses for design calculation.

2.1. The choice of material (Table 1.2). Further presentation will be in parallel: for a spur gear - in the left column, for a helical gear - in the right column.

According to the selected material and surface hardness, contact strength is the main design criterion.

2.2. Permissible fatigue contact stresses of the gear.

The calculation for these allowable stresses prevents fatigue spalling of the working surfaces during a given service life. t.

(6)

where Z R- coefficient taking into account surface roughness (Table 1.3).

Z V- coefficient taking into account peripheral speed. For given values ​​of the shaft speed, it can be preliminarily assumed in what interval the peripheral transmission speed lies (Table 1.3).

S H- safety factor (Table 1.3).

ZN- durability factor

(7)

N HG- base number of cycles

NGH = (HB) 3 £ 12×10 7 . (8)

For a helical gear, if it has HB>350, recalculate units HRC in units HB(Table 1.4).

N HE

N HE 1 = 60x n 1× t× e H. (9)

e H- equivalence factor, which is determined by the loading histogram

, (10)

where Tmax- the largest of the long-acting moments. In our case, this will be the moment T, effective t 1 part of the total operating time t; then q 1 =1.

T i- each subsequent load step acting over time t i =t i × t. The first stage of the histogram, equal in load T peak = q peak × T, is not taken into account when calculating the number of cycles. This load with a small number of cycles has a hardening effect on the surface. It is used to test static strength.

m- the degree of the fatigue curve, equal to 6. Thus,

The equivalence coefficient shows that the moment T operating during e H×t time, has the same fatigue effect as the real load corresponding to the load histogram over time t.

s Hlim- the limit of contact endurance of the gear when the base number of cycles is reached N HG(Table 1.5).

Rated allowable contact stresses for transmission

Point 3. Choice of design coefficients.

3.1 Choice of load factor. The load factor for preliminary calculations is selected from the interval

K H = 1.3...1.5. (16)

If in the calculated gear the gears are located symmetrically with respect to the supports, K H chosen closer to the lower limit. For helical gears K H less is taken due to the greater smoothness of operation and, consequently, less dynamic load.

3.2. Choice of gear width factor (Table 1.6). For gear drives it is recommended:

- for multi-stage y a = 0.315 ... 0.4;

- for single-stage y a = 0.4 ... 0.5;

the upper limit is selected for helical gears;

- for chevron gears y a = 0.630 ... 1.25.

Item 4. Transfer design calculation.

4.1. Determination of center distance.

For a closed gear, if both or at least one of the wheels has a hardness of less than 350 units, the design calculation is carried out for fatigue contact strength to prevent spalling during a given service life t.

, mm. (17)

Here T 1 - moment on the shaft gears in Nm.

Numerical coefficient:

Ka = 450;

Ka= 410.

The calculated center distance is taken as the closest standard according to table 1.7.

4.2. Selecting a normal module. For gear wheels HB£350 for at least one wheel it is recommended to choose a normal module from the following ratio

. (18)

Write out all the standard values ​​of the normal module (Table 1.8) included in the interval (18) .

As a first approximation, one should strive to select the minimum module, however, for power transmissions, a module less than 1.25 mm is not recommended. When choosing a module for a spur gear, in order to avoid modifying the gear, it is necessary that the total number of teeth

turned out to be an integer. Then

If a fractional number is rounded up to a whole number, and the number of teeth of the wheel

4.3. For helical transmission number of teeth

The number of teeth should be rounded to the nearest whole number.

4.5. Pitch diameters

Calculate diameters to the third decimal place.

Run a check

For unmodified transmission and high-altitude modification must be accurate to three decimal places.

4.6. Lug diameters

4.7. Cavity diameters

(26)

4.8. Estimated wheel width

In a split gear, the width of each wheel of the split pair is

In chevron gear full wheel width

where C- the width of the middle groove for the tool exit, is selected from table 1.16. The diameter of the groove is less than the diameter of the cavity by 0.5× m.

4.9. End degree of overlap

. (31)

4.10. Peripheral speed

If the speed differs from that tentatively adopted in paragraph 2.2 when determining the coefficient KV, you should return to clause 2.2 and clarify the allowable stresses.

According to the circumferential speed, select the degree of transmission accuracy (Table 1.9). For gears of general engineering at speeds of not more than 6 m / s for spur gears and not more than 10 m / s for helical gears, the 8th degree of accuracy is selected. The helical gear can be machined to the 7th degree of accuracy, and after surface hardening of the HDTV, the resulting deformations will transfer the gear parameters to the 8th degree of accuracy.

Item 5. Check calculations.

5.1. For verification calculations for both contact and flexural strength, we determine the load factors.

. (33)

. (34)

KHV And KFV- coefficients of internal dynamic load. They are selected from table 1.10. If the speed value falls within the intervals of the range, the coefficient is calculated by interpolation.

KH b And KFb- coefficients of load concentration (uneven load distribution along the length of contact lines). Their values ​​are selected from Table 1.11 by interpolation.

K H a And K F a- coefficients of load distribution between the teeth. Selected from table 1.12 by interpolation.

5.2. Contact voltage test

. (35)

Z E - material coefficient. For steel

Z E = 190.

Z e - coefficient for taking into account the total length of contact lines

Spurs; (36)

Helical; (37)

Z H is the shape factor of the mating surfaces. Selected from table 1.13 by interpolation.

F t- circumferential force

Deviation

. (39)

The sign (+) indicates underload, the sign (-) indicates overload.

RECOMMENDATIONS

Both underload and overload are allowed no more than 5%.

If Ds H goes beyond ±20%, then for the gearbox with standard parameters, the center distance should be changed a W and return to point 4.2.

If Ds H goes beyond ±12%:

In case of underload - reduce y a and return to paragraph 4.8.

In case of overload - increase y a , not exceeding the recommended values ​​for this type of transmission and return to paragraph 4.8. You can change the hardness of the tooth surface within the recommended limits and return to step 2.

If Ds H will be less than 12%, it is possible to correct the allowable stresses by heat treatment and return to point 2.

5.3. Flexural fatigue test.

5.3.1. Permissible bending stresses

. (40)

Testing against these stresses prevents fatigue cracking at the root of the tooth during a given service life. t and consequently, tooth decay.

Y R- coefficient of roughness of the transition curve (Table 1.14).

Y X- scale factor (Table 1.14).

Y d is the coefficient of sensitivity of the material to stress concentration (Table 1.14).

Y A- load reversibility factor (Table 1.14).

Y N- coefficient of durability. Calculated separately for gear and wheel

N FG- base number of cycles. For steel teeth

N FG= 4×10 6 . (42)

m- the degree of fatigue curve. In the previous and subsequent formulas for calculating fatigue flexural strength:

For tempered steels

for hardened steels

NFE 1 - equivalent number of gear cycles

NFE 1 = 60x n 1× t× eF. (43)

eF- equivalence ratio

. (44)

In accordance with the loading histogram, as in the calculation for contact strength,

Equivalent number of wheel cycles

S F ands flim- safety factor and endurance limit of the tooth are selected from table 1.15.

5.3.2. Working stresses of a bend. Determined separately for gear and wheel

. (47)

YFS- tooth shape factor

. (48)

X- tool shift factor.

Z V- equivalent number of teeth

Y e - coefficient taking into account the overlap of the teeth in the mesh

Y b - coefficient of the angle of inclination of the tooth

. (53)

If Y b turned out to be less than 0.7, it should be taken

Y b = 0.7

Working stresses are determined for each gear or for the one with the smaller ratio

Actual flexural fatigue strength

The value of the flexural fatigue safety factor indicates the degree of reliability in relation to the probability of tooth breakage. The higher this coefficient, the lower the probability of tooth fatigue failure.

5.4. Test for contact static strength.

. (56)

Tmax=

[s] Hmax- allowable static contact stresses.

For improved teeth

. (57)

These allowable stresses prevent plastic deformation of the surface layers of the tooth.

The yield strength s T can be selected from table 1.2.

For surface hardened teeth, including hardened HDTV

. (58)

These allowable stresses prevent cracking of the surface layers of the tooth.

5.5. Verification of flexural static strength. Check is done for gear and wheel

. (59)

Permissible static bending stresses. For improved and surface hardened teeth

. (60)

Checking for these allowable stresses prevents instantaneous tooth breakage when the gear is overloaded.

Table 1.1

Table 1.2

Steel grade	Heat treatment	Section size, mm, no more	Surface hardness HB or HRC	Tensile strength s b , MPa	Yield strength s T, MPa
	Improvement		HB 192...228
	Normalization Improvement		HB 170...217 HB 192...217
	Normalization Improvement		HB 179...228 HB 228...255	...800
40X	Improvement Improvement Improvement	100...300 300...500	HB 230...280 HB 163...269 HB 163...269
40HN	Enhance Enhance Temper	100...300	HB 230...300 HB³241 HRC 48...54
20X	Cementation		HRC 56...63
12ХН3А	Cementation		HRC 56...63
38HMYUA	Nitriding	-	HRC 57...67

Note. The size of the section means the radius of the workpiece of the gear shaft or the thickness of the wheel rim.

Table 1.3

Table 1.4

HRC
HB

Table 1.5

Table 1.6

Table 1.8

Table 1.9

Table 1.10

Degree of accuracy	Tooth surface hardness	Transmission type	KHV	KFV
Peripheral speed V, m/s
	HB 1 and HB 2 >350	straight	1,02	1,12	1,25	1,37	1,5	1,02	1,12	1,25	1,37	1,5
scythe	1,01	1,05	1,10	1,15	1,20	1,01	1,05	1,10	1,15	1,20
HB 1 or HB 2 £350	straight	1,04	1,20	1.40	1,60	1,80	1,08	1,40	1,80	-	-
scythe	1,02	1,08	1,16	1,24	1,32	1,03	1,16	1,32	1,48	1,64
	HB 1 and HB 2 >350	straight	1,03	1,15	1,30	1,45	1,60	1,03	1,15	1,30	1,45	1,60
scythe	1,01	1,06	1,12	1,18	1,24	1,01	1,06	1,12	1,18	1,24
HB 1 or HB 2 £350	straight	1,05	1,24	1,48	1,72	1,96	1,10	1,48	1,96	-	-
scythe	1,02	1,10	1,19	1,29	1,38	1,04	1,19	1,38	1,57	1,77
	HB 1 and HB 2 >350	straight	1,03	1,17	1,35	1,52	1,70	1,03	1,17	1,35	1,52	1,70
scythe	1,01	1,07	1,14	1,21	1,28	1,01	1,07	1,14	1,21	1,28
HB 1 or HB 2 £350	straight	1,06	1,28	1,56	1,84	-	1,11	1,56	-	-	-
scythe	1,02	1,11	1,22	1,34	1,45	1,04	1,22	1,45	1,67	-

Table 1.11

Coefficient KH b at HB 1 £350 or HB 2 £350
Transmission design	Coefficient y d = b W/d 1
0,2	0,4	0,6	0,8	1,0	1,2	1,4	1,6	1,8	2,0
Cantilever gear on ball bearings	1,09	1,19	1,3	-	-	-	-	-	-	-
Cantilever gear on roller bearings	1,07	1,13	1,20	1,27	-	-	-	-	-	-
High-speed pair of a two-stage gearbox of an unfolded scheme	1,03	1,06	1,08	1,12	1,16	1,20	1,24	1,29	-	-
Low-speed pair of two-stage coaxial gearbox	1,02	1,03	1,06	1,08	1,10	1,13	1,16	1,19	1,24	1,30
Low-speed pair of a two-stage gearbox of an expanded and coaxial scheme	1,02	1,03	1,04	1,06	1,08	1,10	1,13	1,16	1,19	1,25
Single stage spur gearbox	1,01	1,02	1,02	1,03	1,04	1,06	1,08	1,10	1,14	1,18
Low-speed pair of a two-stage gearbox with a spaced high-speed stage	1,01	1,02	1,02	1,02	1,03	1,04	1,05	1,07	1,08	1,12
Coefficient KFb=(0.8...0.85)× KH b³1

Table 1.12

Table 1.14

Coefficient	Coefficient name	Coefficient value
Y R	Spiral Roughness Coefficient	Gear milling and grinding Y R=1. Polishing Y R=1.05...1.20. Higher values ​​to improve and harden HDTV.
Y X	Size factor (scale factor)	Steel: bulk heat treatment Y X=1.03 - 0.006× m; £0.85 Y X£1. Surface hardening, nitriding Y X=1.05 - 0.005× m; £0.8 Y X£1. Cast iron with spheroidal graphite Y X=1.03 - 0.006× m; £0.85 Y X£1. Gray cast iron Y X=1.075 - 0.01× m;0.7£ Y X£1.
Y d	Coefficient of material sensitivity to stress concentration	Y d = 1.082 - 0.172× lgm.
Continuation of table 1.14
Y A	Reversibility factor	For non-reversible operation Y A=1. In reverse operation with equal loading conditions in both directions: for normalized and tempered steel Y A=0.65; for hardened steel Y A=0.75; for nitrided steel Y A=0,9.

Table 1.15

Heat treatment	Surface hardness	Steel grades	s flim, MPa	S F with the probability of non-destruction
				normal	increased
Normalization, improvement	180...350 HB	40.45,40X, 40XN, 35XM	1.75×( HB)	1,7	2,2
Bulk hardening	45...55 HRC	40H,40HN, 40HFA	500...550	1.7	2,2
HDTV hardening through	48...52 HRC	40X,35XM, 40XN	500...600	1,7	2,2
HDTV surface hardening	48...52 HRC	40X,35XM, 40XN	600...700	1,7	2,2
Nitriding	57...67 HRC	38HMYUA	590...780	1,7	2,2
Cementation	56...63 HRC	12ХН3А	750...800	1,65...1,7	2...2,2

Table 1.16

Module	Helix angle b 0	Module	Tooth angle b 0
m, mm				m, mm
	Groove width C, mm		Groove width C, mm
2,5
3,0
3,5

The required drive power is determined by the formula:

where T 2 – moment on the output shaft (Nm);

n 2 - frequency of rotation of the output shaft (rpm).

Determination of the required power of the electric motor.

The required motor power is determined by the formula

where η gearbox- efficiency of the gearbox;

According to the kinematic scheme of a given drive, the efficiency of the gearbox is determined by the dependence:

η gearbox = η engagement η 2 bearings η couplings ,

where η engagement– gearing efficiency; accept η engagement = 0,97 ;

η bearings– efficiency of a pair of rolling bearings; accept η bearings = 0,99 ;

η couplings– clutch efficiency; accept η couplings = 0,98 .

1.3. Determination of the frequency of rotation of the motor shaft.

We determine the speed range in which the synchronous speed of the electric motor can be located by the formula:

n from = un 2 ,

where u- gear ratio of the stage; we select the range of gear ratios, which is recommended for one stage of a spur gear in the range from 2 - 5.

For example: n from = un 2 = (2 - 5)200 = 400 - 1000 rpm.

1.4. Motor selection.

According to the required power of the electric motor R cons.(given that R el.dv. ≥ R cons.) and synchronous shaft speed n from choose an electric motor:

series…..

power R= ……kW

synchronous speed n from= …..rpm

asynchronous speed n 1 = …..r/min.

Rice. 1. Sketch of the electric motor.

1.5. Determination of the gear ratio of the gearbox.

According to the calculated value of the gear ratio, we select the standard value, taking into account the error, from a series of gear ratios. Accept u Art. = ….. .

1.6. Determination, speeds and torques on the shafts of the gearbox.

Input shaft speed n 1 = ….. rpm.

Output shaft speed n 2 = ….. rpm.

Torque on the output shaft wheel:

Torque on the input shaft gear:

2. CALCULATION OF A CLOSED GEAR.

2.1. Design calculation.

1. Choice of wheel material.

For example:

Gear Wheel

HB = 269…302 HB = 235…262

HB 1 = 285 HB 2 = 250

2. We determine the allowable voltage contacts for the gear teeth and wheels :

where  H lim - endurance limit of the contact surface of the teeth, corresponding to the basic number of cycles of alternating stresses; determined depending on the hardness of the tooth surface or a numerical value is set;

For example:  H lim = 2HB+70.

S H– safety factor; for gears with uniform material structure and tooth surface hardness HB 350 recommended S H = 1,1 ;

Z N– durability coefficient; for gears during long-term operation with a constant load mode, it is recommended Z N = 1 .

Finally, for the allowable contact stress, the smaller of the two values ​​of the allowable contact stresses of the wheel and gear is taken [  H] 2 and [  H ] 1:[ H ] = [ H ] 2 .

3. Determine the center distance from the condition of contact endurance of the active surfaces of the teeth .

where E etc- reduced modulus of elasticity of wheel materials; for steel wheels can be accepted E etc= 210 5 MPa;

 ba- coefficient of wheel width relative to the center distance; for wheels located symmetrically with respect to the supports, it is recommended ψ ba = 0,2 – 0,4 ;

TO H  is the load concentration factor in calculations for contact stresses.

To determine the coefficient TO H  it is necessary to determine the ratio of the relative width of the ring gear relative to the diameter ψ bd : ψ bd = 0,5ψ ba (u1)=….. .

According to the graph of the figure ... .. taking into account the location of the gear relative to the supports, with hardness HB 350, according to the value of the coefficient ψ bd we find: TO H  = ….. .

We calculate the center distance:

For example:

For gearboxes, the center distance is rounded off according to a series of standard center distances or a series Ra 40 .

Assign but W= 120 mm.

4. Determine the transmission module.

m = (0,01 – 0,02)but W= (0.01 - 0.02)120 = 1.2 - 2.4 mm.

For a number of modules from the obtained interval, we assign the standard value of the module: m= 2 mm.

5. Determine the number of gear teeth and wheels.

The total number of teeth of the gear and wheel is determined from the formula: but W = m(z 1 +z 2 )/2;

from here z   = 2but W /m= …..; accept z  = ….. .

Number of gear teeth: z  1 = z  /(u1) = …..

To eliminate undercut teeth z 1 ≥ z min ; for spur engagement z min = 17 . Accept z 1 = ….. .

Number of wheel teeth: z 2 = z  - z 1 = .. Recommended z 2  100 .

6. We specify the gear ratio.

We determine the actual gear ratio by the formula:

The error in the value of the actual gear ratio from the calculated value:

The design accuracy condition is met.

For the gear ratio of the gearbox, we take u fact = ….. .

7. We determine the main geometric dimensions of the gear and wheel.

For wheels cut without tool offset:

pitch circle diameters

d W = d

engagement angle and profile angle

α W = α = 20º

pitch diameters

d 1 = z 1 m

d 2 = z 2 m

tooth tip diameters

d a1 = d 1 +2 m

d a2 = d 2 +2 m

cavity diameters

d f 1 = d 1 –2,5 m

d f 2 = d 2 –2,5 m

tooth height

h = 2,25 m

ring gear width

b w = ψ ba but W

gear and wheel ring width

b 2 = b w

b 1 = b 2 + (3 - 5) = ..... . Accept b 1 = ….. mm.

check the value of the center distance

a w = 0,5  (d 1 + d 2 )