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 (rpm) at the inlet;
N2 - shaft rotation speed (rpm) at the output.
The calculated value is rounded to the value specified in technical characteristics specific type of gearboxes.
Table 2. Range gear ratios for different types of gearboxes
IMPORTANT!
The rotation speed of the electric motor shaft and, accordingly, the gearbox input shaft 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 the 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 the 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 up 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:
- not an easy task. One wrong step in the calculation is fraught not only with premature equipment failure, but also financial losses (especially if the gearbox is in production). Therefore, the calculation of the geared motor is most often entrusted to a specialist. But what to do when you don't have such a specialist?
What is a geared motor for?
Gearmotor - a drive mechanism that is a combination of a gearbox and an electric motor. In this case, the motor is attached to the gearbox on a straight line without special couplings for connection. Due to the high level of efficiency, compact size and ease of maintenance, this type of equipment is used in almost all areas of industry. Gearmotors have found applications in almost all industrial sectors:
How to choose a geared motor?
If the task is to select a geared motor, most often it all comes down to choosing an engine of the required power and the number of revolutions on the output shaft. However, there are other important characteristics that are important to consider when choosing a geared motor:
- Gear motor type
Understanding the type of geared motor can greatly simplify the selection. By the type of transmission, there are: planetary, bevel and coaxial-cylindrical geared motors. They all differ in the arrangement of the shafts.
- Output turns
The rotation speed of the mechanism to which the geared motor is attached is determined by the number of output revolutions. The higher this indicator, the greater the amplitude of rotation. For example, if a geared motor drives a conveyor belt, then the speed of its movement will depend on the speed indicator.
- Electric motor power
The power of the electric motor of the geared motor is determined depending on the required load on the mechanism at a given speed of rotation.
- Features of operation
If you plan to use a geared motor under constant load conditions, when choosing it, be sure to check with the seller for how many hours of continuous operation the equipment is designed for. It will also be important to find out the permissible number of inclusions. Thus, you will know exactly after what period of time you will have to replace the equipment.
Important: The service life of high-quality geared motors with active 24/7 operation should be at least 1 year (8760 hours).
- Working conditions
Before ordering a geared motor, it is necessary to determine its location and operating conditions of the equipment (indoors, under a canopy or in the open air). This will help you set a clearer task for the seller, and for him, in turn, select a product that clearly meets your requirements. For example, special oils are used to facilitate the operation of a geared motor at very low or very high temperatures.
How to calculate a geared motor?
Mathematical formulas are used to calculate all the necessary characteristics of the geared motor. Determining the type of equipment also largely depends on what it will be used for: for lifting mechanisms, mixing or for moving mechanisms. So for lifting equipment, worm and 2MCH gear motors are most often used. In such gearboxes, the possibility of rotating the output shaft when a force is applied to it is excluded, which eliminates the need to install a shoe brake on the mechanism. For various mixing mechanisms, as well as for various drilling rigs, gearboxes of the 3MP (4MP) type are used, since they are able to evenly distribute the radial load. If high torque values are required in movement mechanisms, gear motors of the 1MTs2S, 4MTs2S types are most often used.
Calculation of the main indicators for choosing a geared motor:
- Calculation of the revolutions at the output of the geared motor.
The calculation is made according to the formula:
V = ∏ * 2R * n \ 60
R - radius of the lifting drum, m
V - lifting speed, m * min
n - revolutions at the output of the geared motor, rpm
- Determination of the angular speed of rotation of the gear motor shaft.
The calculation is made according to the formula:
ω = ∏ * n \ 30
- Torque calculation
The calculation is made according to the formula:
M = F * R (H * M)
Important: The rotation speed of the electric motor shaft and, accordingly, the gearbox input shaft 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. Manufacturers indicate this technical parameter in the summary characteristics of electric motors.
- Determination of the required power of the electric motor
The calculation is made according to the formula:
P = ω * M, W
Important:Correctly calculated drive power helps to overcome the mechanical frictional resistance that occurs during straight and rotary movements. If the power exceeds the required by more than 20%, this will complicate the control of the shaft speed and adjust it to the required value.
Where to buy a geared motor?
It is not difficult to buy today. The market is overflowing with offers from various manufacturing plants and their representatives. Most of the manufacturers have their own online store or official website on the Internet.
When choosing a supplier, try to compare not only the price and characteristics of geared motors, but also check the company itself. The presence of letters of recommendation, certified by the seal and signature from customers, as well as qualified specialists in the company, will help protect you not only from additional financial costs, but also secure the operation of your production.
Having problems with the selection of a geared motor? Contact our specialists for help by contacting us by phone or leaving a question to the author of the article.
Course work
Discipline Machine parts
Theme "Calculation of the gearbox"
Introduction
1. Kinematic diagram and initial data
2. Kinematic calculation and selection of an electric motor
3. Calculation of gears of the reducer
4. Preliminary calculation of gear shafts and selection of bearings
5. Structural dimensions of gear and wheel
6. Constructive dimensions of the gearbox housing
7. The first stage of the gearbox layout
8. Checking the durability of the bearing
9. The second stage of the layout. Checking the strength of keyed connections
10. Revised calculation of shafts
11. Drawing the gearbox
12. Landing gear, cogwheel, bearing
13. Choice of oil grade
14. Assembling the gearbox
Introduction
A gearbox is a mechanism consisting of gear or worm gears, made in the form of a separate unit and serving to transfer rotation from the engine shaft to the shaft of the working machine. The kinematic diagram of the drive can include, in addition to the gearbox, open gear drives, chain or belt drives. These mechanisms are the most common topic of course design.
The purpose of the gearbox is to lower the angular velocity and, accordingly, to increase the torque of the driven shaft compared to the driving one. The mechanisms for increasing the angular velocity, made in the form of separate units, are called accelerators or multipliers.
The gearbox consists of a body (cast iron or welded steel), in which transmission elements are placed - gears, shafts, bearings, etc. a gear oil pump) or cooling devices (for example, a cooling water coil in a worm gear housing).
The gearbox is designed either for the drive of a specific machine, or for a given load (torque on the output shaft) and gear ratio without specifying a specific purpose. The second case is typical for specialized factories that organize the serial production of gearboxes.
Kinematic diagrams and general views of the most common types of gearboxes are shown in Fig. 2.1-2.20 [L.1]. On the kinematic diagrams, the letter B denotes the input (high-speed) shaft of the gearbox, the letter T - the output (low-speed) shaft.
Gearboxes are classified according to the following main features: type of transmission (gear, worm or gear-worm); the number of stages (one-stage, two-stage, etc.); type - gears (cylindrical, bevel, bevel-cylindrical, etc.); the relative position of the gearbox shafts in space (horizontal, vertical); the peculiarities of the kinematic scheme (expanded, coaxial, with a bifurcated stage, etc.).
Possibilities of obtaining large gear ratios with small dimensions are provided by planetary and wave gearboxes.
1. Kinematic diagram of the gearbox
Initial data:
Conveyor drive shaft power
;Reducer shaft angular speed
;Gear ratio of the reducer
;Deviation from gear ratio
;Gearbox running time
.1 - electric motor;
2 - belt drive;
3 - elastic sleeve-finger coupling;
4 - reducer;
5 - belt conveyor;
I - electric motor shaft;
II - the drive shaft of the gearbox;
III - the driven shaft of the gearbox.
2. Kinematic calculation and selection of an electric motor
2.1 According to table 1.1 efficiency of a pair of cylindrical gears η 1 = 0.98; coefficient taking into account the loss of a pair of rolling bearings, η 2 = 0.99; V-belt transmission efficiency η 3 = 0.95; Efficiency of a flat-belt transmission in the bearings of the drive drum, η 4 = 0.99
2.2 General drive efficiency
η = η 1 η2 η 3 η 4 = 0.98 ∙ 0.99 2 ∙ 0.95 ∙ 0.99 = 0.90
2.3 Required motor power
= = 1.88 kW.where P III is the power of the drive output shaft,
h is the overall efficiency of the drive.
2.4 According to GOST 19523-81 (see table P1 in the appendices [L.1]), according to the required power P dv = 1.88 kW, we select a three-phase asynchronous squirrel-cage motor of the 4A series closed, blown, with a synchronous speed of 750 rpm 4A112MA8 with parameters P dv = 2.2kW and slip 6.0%.
Rated speed
n dv. = n c (1-s)
where n c is the synchronous rotation frequency,
s- slip
2.5 Angular velocity
= = 73.79 rad / s.2.6 Speed
= = 114.64rpm2.7 Gear Ratio
= = 6,1where w I is the angular speed of the engine,
w III - angular speed of the output drive
2.8 We plan for the gearbox u = 1.6; then for V-belt transmission
= = 3.81 - what is within the recommended2.9 Torque generated on each shaft.
kN × m.Torque on the 1st shaft M I = 0.025kN × m.
P II = P I × h p = 1.88 × 0.95 = 1.786 N × m.
rad / s kN × m.Torque on the 2nd shaft M II = 0.092 kN × m.
kN × m.Torque on the 3rd shaft M III = 0.14 kN × m.
2.10 Let's check:
Determine the speed on the 2nd shaft:
Rotational frequencies and angular speeds of shafts
3. Calculation of gears of the reducer
We select materials for gears the same as in § 12.1 [L.1].
For gear, steel 45, heat treatment - improvement, hardness HB 260; for wheel steel 45, heat treatment - improvement, hardness HB 230.
The permissible contact stress for spur gears made of the indicated materials is determined using the formula 3.9, p. 33:
where s H limb is the limit of contact endurance; For a wheel
= MPa.Acceptable contact voltage
= 442 MPa.I accept the ratio of the crown width ψ bRe = 0.285 (according to GOST 12289-76).
The coefficient K nβ, taking into account the uneven distribution of the load along the width of the crown, is taken according to table. 3.1 [L.1]. Despite the symmetrical arrangement of the wheels relative to the supports, we will accept the value of this coefficient, as in the case of an asymmetric arrangement of the wheels, since a pressure force acts on the drive shaft from the side of the V-belt transmission, causing its deformation and worsening the contact of the teeth: K nβ = 1.25.
In this formula for spur gears K d = 99;
Gear ratio U = 1.16;
M III - torque on the 3rd shaft.
The design engineer is the creator of new technology, and the rate of scientific and technological progress is largely determined by the level of his creative work. The activity of a designer is one of the most complex manifestations of the human mind. The decisive role of success in the creation of new technology is determined by what is laid down in the designer's drawing. With the development of science and technology, problematic issues are resolved taking into account an increasing number of factors based on data from various sciences. During the implementation of the project, mathematical models are used based on theoretical and experimental studies related to volumetric and contact strength, materials science, heat engineering, hydraulics, elasticity theory, and structural mechanics. Information from courses on strength of materials, theoretical mechanics, mechanical engineering, etc. is widely used. All this contributes to the development of independence and a creative approach to the problems posed.
When choosing the type of gearbox for the drive of the working body (device), it is necessary to take into account many factors, the most important of which are: the value and nature of the load change, the required durability, reliability, efficiency, weight and dimensions, noise requirements, product cost, operating costs.
Of all types of gears, gears have the smallest dimensions, weight, cost and friction losses. The loss factor of one gear pair, if carefully executed and properly lubricated, does not usually exceed 0.01. Gear drives, in comparison with other mechanical transmissions, have great reliability in operation, a constant gear ratio due to the absence of slippage, the ability to be used in a wide range of speeds and gear ratios. These properties ensured a wide distribution of gears; they are used for capacities ranging from negligible (in devices) to tens of thousands of kilowatts.
The disadvantages of gears include the requirements for high manufacturing accuracy and noise when working at significant speeds.
Helical gears are used for critical transmissions at medium and high speeds. The volume of their application is over 30% of the volume of application of all cylindrical wheels in machines; and this percentage is constantly increasing. Helical gears with hard tooth surfaces require increased protection against contamination to avoid uneven wear along the length of the contact lines and the risk of chipping.
One of the goals of the completed project is the development of engineering thinking, including the ability to use previous experience, to model using analogs. For a course project, objects are preferred that are not only well-distributed and of great practical importance, but also are not subject to obsolescence in the foreseeable future.
There are various types of mechanical transmissions: cylindrical and bevel, with straight teeth and helical, hypoid, worm, globoid, single and multi-threaded, etc. This raises the question of choosing the most rational transmission option. When choosing the type of transmission, they are guided by indicators, among which the main ones are efficiency, overall dimensions, weight, smooth operation and vibration load, technological requirements, and the preferred number of products.
When choosing the types of gears, the type of engagement, the mechanical characteristics of materials, it is necessary to take into account that the costs of materials make up a significant part of the cost of the product: in general-purpose gearboxes - 85%, in road cars - 75%, in cars - 10%, etc.
The search for ways to reduce the mass of projected objects is the most important prerequisite for further progress, a prerequisite for the conservation of natural resources. Most of the currently generated energy is accounted for by mechanical transmissions, so their efficiency to a certain extent determines the operating costs.
The drive with the use of an electric motor and a gearbox with external gear meets the requirements to reduce weight and overall dimensions to the fullest extent.
Electric motor selection and kinematic calculation
According to the table. 1.1 we will accept the following values of efficiency:
- for a closed spur gear transmission: h1 = 0.975
- for a closed spur gear transmission: h2 = 0.975
The overall efficiency of the drive will be:
h = h1 ·… · hn · h 3 h couplings2 = 0.975 0.975 0.993 0.982 = 0.886
where hsubsh. = 0.99 - efficiency of one bearing.
hclutch = 0.98 - efficiency of one clutch.
The angular velocity at the output shaft will be:
wout. = 2 V / D = 2 3 103/320 = 18.75 rad / s
The required engine power will be:
Preq. = F V / h = 3.5 3 / 0.886 = 11.851 kW
In table P. 1 (see Appendix), according to the required power, we select the 160S4 electric motor, with a synchronous speed of 1500 rpm, with the parameters: Pmotor = 15 kW and slip 2.3% (GOST 19523–81). Rated speed neng. = 1500–1500 · 2.3 / 100 = 1465.5 rpm, angular speed w = p n engine / 30 = 3.14 1465.5 / 30 = 153.467 rad / s.
Overall gear ratio:
u = w input. / wout. = 153.467 / 18.75 = 8.185
The following gear ratios were chosen for the gears:
The calculated frequencies and angular speeds of rotation of the shafts are summarized in the table below:
Shaft power:
P1 = Preq. · Hb. H (couplings 1) = 11.851 103 0.99 0.98 = 11497.84 W
P2 = P1 h1 hsupport = 11497.84 0.975 0.99 = 11098.29 W
P3 = P2 h2 h bearing = 11098.29 0.975 0.99 = 10393.388 W
Torques on shafts:
T1 = P1 / w1 = (11497.84 · 103) / 153.467 = 74,920.602 N mm
T2 = P2 / w2 = (11098.29 103) / 48.72 = 227797.414 N mm
T3 = P3 / w3 = (10393.388 103) / 19.488 = 533322.455 N mm
According to table P. 1 (see the Appendix of Chernavsky's textbook), the 160S4 electric motor is selected, with a synchronous speed of 1500 rpm, with a power of Pmotor = 15 kW and a slip of 2.3% (GOST 19523–81). Nominal rotational speed considering slip nmotor = 1465.5 rpm.
Gear ratios and gear efficiency
Calculated frequencies, angular speeds of rotation of shafts and moments on shafts
2. Calculation of the 1st spur gear
Hub diameter: dstop = (1.5 ... 1.8) · dshaft = 1.5 · 50 = 75 mm.
Hub length: Lstup = (0.8 ... 1.5) · dshaft = 0.8 · 50 = 40 mm = 50 mm.
5.4 Cylindrical gear 2nd gear
Hub diameter: dstop = (1.5 ... 1.8) · dshaft = 1.5 · 65 = 97.5 mm. = 98 mm.
Hub length: Lstup = (0.8 ... 1.5) · dshaft = 1 · 65 = 65 mm
Rim thickness: dо = (2.5 ... 4) · mn = 2.5 · 2 = 5 mm.
Since the thickness of the rim must be at least 8 mm, then we take dо = 8 mm.
where mn = 2 mm is the normal modulus.
Disc thickness: С = (0.2 ... 0.3) · b2 = 0.2 · 45 = 9 mm
where b2 = 45 mm is the width of the ring gear.
Rib thickness: s = 0.8 C = 0.8 9 = 7.2 mm = 7 mm.
Inner rim diameter:
Doboda = Da2 - 2 (2 mn + do) = 262 - 2 (2 2 + 8) = 238 mm
Center Circle Diameter:
DC hole = 0.5 (Doboda + dstep.) = 0.5 (238 + 98) = 168 mm = 169 mm
where Doboda = 238 mm is the inner diameter of the rim.
Hole diameter: D = Doboda - dstep.) / 4 = (238 - 98) / 4 = 35 mm
Chamfer: n = 0.5 mn = 0.5 2 = 1 mm
6. Choice of couplings
6.1 Selection of the coupling on the input shaft of the drive
Since there is no need for large compensating capacities of the couplings and, during installation and operation, sufficient alignment of the shafts is observed, it is possible to select an elastic coupling with a rubber star. Couplings have high radial, angular and axial stiffness. The choice of an elastic coupling with a rubber sprocket is made depending on the diameters of the shafts to be connected, the calculated transmitted torque and the maximum permissible shaft speed. Connected shafts diameters:
d (electric motor) = 42 mm;
d (1st shaft) = 36 mm;
Transmitted torque through the clutch:
T = 74.921 Nm
Estimated transmittable torque through the coupling:
Tr = kr · T = 1.5 · 74.921 = 112.381 N · m
here kр = 1.5 is a coefficient that takes into account the operating conditions; its values are given in table 11.3.
Coupling speed:
n = 1465.5 rpm
We select an elastic clutch with a rubber sprocket 250–42–1–36–1-U3 GOST 14084–93 (according to table K23) For a design torque of more than 16 Nm, the number of “rays” of the asterisk will be 6.
The radial force with which the elastic coupling with an asterisk acts on the shaft is equal to:
Fm = СDr · Dr,
where: СDr = 1320 N / mm is the radial stiffness of this coupling; Dr = 0.4 mm - radial displacement. Then:
Torque on the shaft Tcr. = 227 797.414 H mm.
2 section
The shaft diameter in this section is D = 50 mm. The stress concentration is due to the presence of two keyways. The width of the keyway is b = 14 mm, the depth of the keyway is t1 = 5.5 mm.
sv = Mizg. / Wnet = 256626.659 / 9222.261 = 27.827 MPa,
3.142 503/32 - 14 5.5 (50 - 5.5) 2/50 = 9222.261 mm 3,
sm = Fa / (p D2 / 4) = 0 / (3.142 502/4) = 0 MPa, Fa = 0 MPa - longitudinal force,
- ys = 0.2 - see page 164;
- es = 0.85 - found according to table 8.8;
Ss = 335.4 / ((1.8 / (0.85 0.97)) 27.827 + 0.2 0) = 5.521.
tv = tm = tmax / 2 = 0.5 Tcr. / Wk net = 0.5 227797.414 / 21494.108 = 5.299 MPa,
3.142 503/16 - 14 5.5 (50 - 5.5) 2/50 = 21494.108 mm 3,
where b = 14 mm is the width of the keyway; t1 = 5.5 mm is the depth of the keyway;
- yt = 0.1 - see page 166;
- et = 0.73 - found according to table 8.8;
St = 194.532 / ((1.7 / (0.73 0.97)) 5.299 + 0.1 5.299) = 14.68.
S = Ss St / (Ss2 + St2) 1/2 = 5.521 14.68 / (5.5212 + 14.682) 1/2 = 5.168
3 section
The shaft diameter in this section is D = 55 mm. The stress concentration is due to the presence of two keyways. Keyway width b = 16 mm, keyway depth t1 = 6 mm.
Safety factor for normal stresses:
Ss = s-1 / ((ks / (es b)) sv + ys sm), where:
- amplitude of the cycle of normal stresses:
sv = Mizg. / Wnet = 187629.063 / 12142.991 = 15.452 MPa,
Wnet = p D3 / 32 - b t1 (D - t1) 2 / D =
3.142 553/32 - 16 6 (55 - 6) 2/55 = 12 142.991 mm 3,
- average stress of the cycle of normal stresses:
sm = Fa / (p D2 / 4) = 0 / (3.142 552/4) = 0 MPa, Fa = 0 MPa - longitudinal force,
- ys = 0.2 - see page 164;
- b = 0.97 - coefficient taking into account the surface roughness, see page 162;
- ks = 1.8 - found according to table 8.5;
Ss = 335.4 / ((1.8 / (0.82 0.97)) 15.452 + 0.2 0) = 9.592.
Safety factor for shear stresses:
St = t-1 / ((k t / (et b)) tv + yt tm), where:
- amplitude and average voltage of the zero cycle:
tv = tm = tmax / 2 = 0.5 Tcr. / Wk net = 0.5 227797.414 / 28476.818 = 4 MPa,
Wk net = p D3 / 16 - b t1 (D - t1) 2 / D =
3.142 553/16 - 16 6 6 (55 - 6) 2/55 = 28476.818 mm 3,
where b = 16 mm is the width of the keyway; t1 = 6 mm is the depth of the keyway;
- yt = 0.1 - see page 166;
- b = 0.97 - coefficient taking into account surface roughness, see page 162.
- kt = 1.7 - found according to table 8.5;
St = 194.532 / ((1.7 / (0.7 0.97)) 4 + 0.1 4) = 18.679.
Resulting safety factor:
S = Ss St / (Ss2 + St2) 1/2 = 9.592 18.679 / (9.5922 + 18.6792) 1/2 = 8.533
The calculated value turned out to be more than the minimum allowable [S] = 2.5. The section is in terms of strength.
12.3 Calculation of the 3rd shaft
Torque on the shaft Tcr. = 533322.455 Hmm.
The material selected for this shaft is steel 45. For this material:
- ultimate strength sb = 780 MPa;
- the endurance limit of steel at a symmetric bending cycle
s-1 = 0.43 sb = 0.43 780 = 335.4 MPa;
- the endurance limit of steel at a symmetrical torsion cycle
t-1 = 0.58 s-1 = 0.58 335.4 = 194.532 MPa.
1 section
The shaft diameter in this section is D = 55 mm. This section, when transmitting torque through the clutch, is calculated for torsion. The stress concentration is caused by the presence of the keyway.
Safety factor for shear stresses:
St = t-1 / ((k t / (et b)) tv + yt tm), where:
- amplitude and average voltage of the zero cycle:
tv = tm = tmax / 2 = 0.5 Tcr. / Wk net = 0.5 533322.455 / 30572.237 = 8.722 MPa,
Wk net = p D3 / 16 - b t1 (D - t1) 2 / (2 D) =
3.142 553/16 - 16 6 (55 - 6) 2 / (2 55) = 30572.237 mm 3
where b = 16 mm is the width of the keyway; t1 = 6 mm is the depth of the keyway;
- yt = 0.1 - see page 166;
- b = 0.97 - coefficient taking into account surface roughness, see page 162.
- kt = 1.7 - found according to table 8.5;
- et = 0.7 - we find it according to table 8.8;
St = 194.532 / ((1.7 / (0.7 0.97)) 8.722 + 0.1 8.722) = 8.566.
The radial force of the coupling acting on the shaft is found in the section "Selecting couplings" and is equal to F couplings. = 225 N
Mizg. = T coupling. L / 2 = 2160 225/2 = 243000 N mm.
Safety factor for normal stresses:
Ss = s-1 / ((ks / (es b)) sv + ys sm), where:
- amplitude of the cycle of normal stresses:
sv = Mizg. / Wnet = 73028.93 / 14238.409 = 17.067 MPa,
Wnet = p D3 / 32 - b t1 (D - t1) 2 / (2 D) =
3.142 553/32 - 16 6 (55 - 6) 2 / (2 55) = 14238.409 mm 3,
where b = 16 mm is the width of the keyway; t1 = 6 mm is the depth of the keyway;
- average stress of the cycle of normal stresses:
sm = Fa / (p D2 / 4) = 0 / (3.142 552/4) = 0 MPa, where
Fa = 0 MPa - longitudinal force in the section,
- ys = 0.2 - see page 164;
- b = 0.97 - coefficient taking into account the surface roughness, see page 162;
- ks = 1.8 - found according to table 8.5;
- es = 0.82 - found according to table 8.8;
Ss = 335.4 / ((1.8 / (0.82 0.97)) 17.067 + 0.2 0) = 8.684.
Resulting safety factor:
S = Ss St / (Ss2 + St2) 1/2 = 8.684 8.566 / (8.6842 + 8.5662) 1/2 = 6.098
The calculated value turned out to be more than the minimum allowable [S] = 2.5. The section is in terms of strength.
2 section
The shaft diameter in this section is D = 60 mm. The stress concentration is due to the bearing fit with guaranteed interference (see table 8.7).
Safety factor for normal stresses:
Ss = s-1 / ((ks / (es b)) sv + ys sm), where:
- amplitude of the cycle of normal stresses:
sv = Mizg. / Wnet = 280800 / 21205.75 = 13.242 MPa,
Wnet = p D3 / 32 = 3.142 603/32 = 21 205.75 mm 3
- average stress of the cycle of normal stresses:
sm = Fa / (p D2 / 4) = 0 / (3.142 602/4) = 0 MPa, Fa = 0 MPa - longitudinal force,
- ys = 0.2 - see page 164;
- b = 0.97 - coefficient taking into account the surface roughness, see page 162;
- ks / es = 3.102 - found according to table 8.7;
Ss = 335.4 / ((3.102 / 0.97) 13.242 + 0.2 0) = 7.92.
Safety factor for shear stresses:
St = t-1 / ((k t / (et b)) tv + yt tm), where:
- amplitude and average voltage of the zero cycle:
tv = tm = tmax / 2 = 0.5 Tcr. / Wk net = 0.5 533322.455 / 42411.501 = 6.287 MPa,
Wk net = p D3 / 16 = 3.142 603/16 = 42411.501 mm 3
- yt = 0.1 - see page 166;
- b = 0.97 - coefficient taking into account surface roughness, see page 162.
- kt / et = 2.202 - found according to table 8.7;
St = 194.532 / ((2.202 / 0.97) 6.287 + 0.1 6.287) = 13.055.
Resulting safety factor:
S = Ss St / (Ss2 + St2) 1/2 = 7.92 13.055 / (7.922 + 13.0552) 1/2 = 6.771
The calculated value turned out to be more than the minimum allowable [S] = 2.5. The section is in terms of strength.
3 section
The shaft diameter in this section is D = 65 mm. The stress concentration is due to the presence of two keyways. Keyway width b = 18 mm, keyway depth t1 = 7 mm.
Safety factor for normal stresses:
Ss = s-1 / ((ks / (es b)) sv + ys sm), where:
- amplitude of the cycle of normal stresses:
sv = Mizg. / Wnet = 392181.848 / 20440.262 = 19.187 MPa,
Wnet = p D3 / 32 - b t1 (D - t1) 2 / D = 3.142 653/32 - 18 7 (65 - 7) 2/65 = 20440.262 mm 3,
- average stress of the cycle of normal stresses:
sm = Fa / (p D2 / 4) = 0 / (3.142 652/4) = 0 MPa, Fa = 0 MPa - longitudinal force,
- ys = 0.2 - see page 164;
- b = 0.97 - coefficient taking into account the surface roughness, see page 162;
- ks = 1.8 - found according to table 8.5;
- es = 0.82 - found according to table 8.8;
Ss = 335.4 / ((1.8 / (0.82 0.97)) 19.187 + 0.2 0) = 7.724.
Safety factor for shear stresses:
St = t-1 / ((k t / (et b)) tv + yt tm), where:
- amplitude and average voltage of the zero cycle:
tv = tm = tmax / 2 = 0.5 Tcr. / Wk net = 0.5 533322.455 / 47401.508 = 5.626 MPa,
Wk net = p D3 / 16 - b t1 (D - t1) 2 / D =
3.142 653/16 - 18 7 (65 - 7) 2/65 = 47401.508 mm 3,
where b = 18 mm is the width of the keyway; t1 = 7 mm is the depth of the keyway;
- yt = 0.1 - see page 166;
- b = 0.97 - coefficient taking into account surface roughness, see page 162.
- kt = 1.7 - found according to table 8.5;
- et = 0.7 - we find it according to table 8.8;
St = 194.532 / ((1.7 / (0.7 0.97)) 5.626 + 0.1 5.626) = 13.28.
Resulting safety factor:
S = Ss St / (Ss2 + St2) 1/2 = 7.724 13.28 / (7.7242 + 13.282) 1/2 = 6.677
The calculated value turned out to be more than the minimum allowable [S] = 2.5. The section is in terms of strength.
13. Thermal design of the gearbox
For the designed gearbox, the area of the heat-dissipating surface is A = 0.73 mm 2 (here the bottom area was also taken into account, because the design of the support legs ensures air circulation around the bottom).
According to the formula 10.1, the condition of the gearbox operation without overheating during continuous operation:
Dt = tm - tv = Ptr (1 - h) / (Kt A) £,
where Rtr = 11.851 kW is the required power for the operation of the drive; tm - oil temperature; tv - air temperature.
We assume that normal air circulation is ensured, and we take the heat transfer coefficient Kt = 15 W / (m2 oC). Then:
Dt = 11851 (1 - 0.886) / (15 0.73) = 123.38o>,
where = 50oС is the permissible temperature difference.
To decrease Dt, the heat-transfer surface of the gearbox housing should be increased accordingly in proportion to the ratio:
Dt / = 123.38 / 50 = 2.468, making the body ribbed.
14. Choice of oil grade
Lubrication of the gearbox elements is carried out by dipping the lower elements in oil, poured into the housing to a level that ensures the immersion of the gearbox by about 10–20 mm. The volume of the oil bath V is determined at the rate of 0.25 dm3 of oil per 1 kW of transmitted power:
V = 0.25 11.851 = 2.963 dm3.
According to table 10.8, we set the oil viscosity. At contact stresses sH = 515.268 MPa and speed v = 2.485 m / s, the recommended oil viscosity should be approximately equal to 30 · 10–6 m / s2. According to table 10.10, we accept industrial oil I-30A (according to GOST 20799-75 *).
We select UT-1 grease for rolling bearings in accordance with GOST 1957–73 (see Table 9.14). The bearing chambers are filled with this grease and replenished periodically.
15. Choice of landings
Landing of gear elements on shafts - Н7 / р6, which according to ST SEV 144–75 corresponds to an easy-press fit.
Landing of couplings on gear shafts - Н8 / h8.
Shaft journals for bearings are made with shaft deflection k6.
The rest of the landings are assigned using the data in Table 8.11.
16. Technology of assembly of the reducer
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 general drawing of the gearbox, starting with the shaft assemblies.
The keys are laid on the shafts and the gear elements of the gearbox are pressed on. Grease rings and bearings should be fitted, preheated in oil to 80-100 degrees Celsius, in series with the gear elements. The assembled shafts are placed in the base of the gearbox housing and put on the housing cover, pre-covering the joint surfaces of 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, bearing caps with a set of metal gaskets are placed, and the thermal gap is adjusted. Before setting the through covers, felt seals soaked in hot oil are placed in the grooves. Check by turning the shafts that the bearings are not jammed (the shafts must be turned by hand) and fix the cover with screws. Then screw in the oil drain plug with a gasket and a rod oil indicator. Pour oil into the body and close the inspection hole with a cover with a gasket, 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.
Conclusion
During the course project on "Machine parts", the knowledge gained over the past period of study in such disciplines as: theoretical mechanics, strength of materials, materials science was consolidated.
The aim of this project is to design a chain conveyor drive, which consists of both simple standard parts and parts, the shape and dimensions of which are determined on the basis of design, technological, economic and other standards.
In the course of solving the problem posed to me, I mastered the method of selecting drive elements, acquired design skills that allow me to provide the required technical level, reliability and long service life of the mechanism.
The experience and skills gained during the course of the course project will be in demand in the implementation of both course projects and the graduation project.
It can be noted that the designed gearbox has good properties in all respects.
According to the results of the calculation for contact endurance, the effective stresses in the engagement are less than the permissible stresses.
According to the calculation results for bending stresses, the effective bending stresses are less than the permissible stresses.
The calculation of the shaft showed that the margin of safety is greater than the allowable one.
The required dynamic load carrying capacity of rolling bearings is less than the rated one.
When calculating, an electric motor was selected that meets the specified requirements.
List of used literature
1. Chernavsky S.A., Bokov K.N., Chernin I.M., Itskevich G.M., Kozintsov V.P. "Course Design of Machine Parts": A study guide for students. M.: Mechanical Engineering, 1988, 416 p.
2. Dunaev P.F., Lelikov O.P. "Design of units and parts of machines", M .: Publishing center "Academy", 2003, 496 p.
3. Sheinblit A.E. "Course design of machine parts": Textbook, ed. 2nd revised and add. - Kaliningrad: "Amber Skaz", 2004, 454 p .: ill., Devil. - B.ts.
4. Berezovsky Yu.N., Chernilevsky D.V., Petrov M.S. "Machine parts", M .: Mechanical engineering, 1983, 384 p.
5. Bokov V.N., Chernilevsky D.V., Budko P.P. "Machine parts: Atlas of structures. M .: Mechanical engineering, 1983, 575 p.
6. Guzenkov PG, "Machine parts". 4th ed. M .: Higher school, 1986, 360 p.
7. Machine parts: Atlas of structures / Ed. D.R. Reshetova. Moscow: Mechanical Engineering, 1979, 367 p.
8. Druzhinin N.S., Tsylbov P.P. Execution of drawings according to ESKD. M .: Publishing house of standards, 1975, 542 p.
9. Kuzmin A.V., Chernin I.M., Kozintsov B.P. "Calculations of machine parts", 3rd ed. - Minsk: Higher school, 1986, 402 p.
10. Kuklin NG, Kuklina GS, "Machine parts" 3rd ed. M .: Higher school, 1984, 310 p.
11. "Gearmotors and Gear Units": Catalog. M .: Publishing house of standards, 1978, 311 p.
12. Perel L. Ya. "Rolling bearings". M .: Mechanical engineering, 1983, 588 p.
13. "Rolling bearings": Directory-catalog / Ed. R.V. Korostashevsky and V.N. Naryshkina. Moscow: Mechanical Engineering, 1984, 280 p.
Program description
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The program is written in Exsel, very easy to use and learn. The calculation is made according to the Chernaski method.
1. Initial data:
1.1. Allowable contact voltage, Mpa;
1.2. The adopted gear ratio, U;
1.3. Torque on the pinion shaft t1, kN * mm;
1.4. Torque on the wheel shaft t2, kN * mm;
1.5. Coefficient;
1.6. The ratio of the width of the crown at the center distance.
2. Standard circumferential module, mm:
2.1. permissible min;
2.2. Allowable max;
2.3 Accepted in accordance with GOST.
3. Calculation of the number of teeth:
3.1. The adopted gear ratio, u;
3.2. Accepted center distance, mm;
3.3. Adopted meshing module;
3.4. Number of gear teeth (accepted);
3.5. Number of wheel teeth (accepted).
4. Calculation of wheel diameters;
4.1. Calculation of pitch diameters of gears and wheels, mm;
4.2. Calculation of the diameters of the tops of the teeth, mm.
5. Calculation of other parameters:
5.1. Calculation of the width of the gear and wheel, mm;
5.2. The peripheral speed of the gear.
6. Checking contact voltages;
6.1. Calculation of contact stresses, MPa;
6.2. Comparison with the permissible contact voltage.
7. Forces in engagement;
7.1. Calculation of the circumferential force, N;
7.2. Calculation of the radial force, N;
7.3. Equivalent number of teeth;
8. Bending stress:
8.1. Choice of gear and wheel material;
8.2. Calculation of the allowable voltage
9. Bending stress check;
9.1. Calculation of bending stress of gear and wheel;
9.2. The fulfillment of the conditions.
Spur gear is the most common mechanical transmission with direct contact. Spur gear is less durable than others and less durable. In such a transmission, during operation, only one tooth is loaded, and vibration is also created during the operation of the mechanism. Due to this, it is impossible and impractical to use such a transmission at high speeds. The service life of a spur gear train is much lower than that of other gear drives (helical, chevron, curved, etc.). The main advantages of such a transmission are the ease of manufacture and the absence of axial force in the bearings, which reduces the complexity of the gearbox bearings, and, accordingly, reduces the cost of the gearbox itself.
