Urgent! ~ Mechanical Design Course Design---Reducer (Everyone, do yourself a favor, good answers will be rewarded!!!!)

For your reference

1. Preface

(1)

Design purpose:

Passed This course is designed to comprehensively apply the basic theoretical knowledge learned, cultivate structural design and calculation skills, and become familiar with the general mechanical device design process.

(2)

Analysis of transmission scheme

The machine is generally composed of a prime mover, a transmission device and a working device. The transmission device is used to transmit the motion and power of the prime mover and transform its motion form to meet the needs of the working device. It is an important part of the machine. Whether the transmission device is reasonable will directly affect the working performance, weight and cost of the machine. In addition to meeting the functions of the working device, a reasonable transmission scheme also requires simple structure, convenient manufacturing, low cost, high transmission efficiency and easy use and maintenance.

In this design, the prime mover is an electric motor and the working machine is a belt conveyor. The transmission scheme adopts a two-stage transmission. The first-stage transmission is a belt drive, and the second-stage transmission is a single-stage spur gear reducer.

Belt transmission has a low load-bearing capacity. When transmitting the same torque, the structural size is larger than other forms. However, it has the advantage of overload protection and can also alleviate impact and vibration, so it is arranged at the high-speed level of the transmission. To reduce the transmitted torque and reduce the structural size of the belt drive.

Gear transmission has high transmission efficiency, wide applicable power and speed range, and long service life. It is one of the most widely used mechanisms in modern machines. This design uses a single-stage spur gear transmission.

The reducer box adopts a horizontally split structure and is cast from HT200 gray cast iron.

2. Parameter design of the transmission system

Original data: working tension of the conveyor belt F=0.2 KN; belt speed V=2.0m/s; drum diameter D=400mm (drum The efficiency is 0.96).

Working conditions: The expected service life is 8 years, the work is two shifts, and the load is light.

Working environment: There is a lot of dust in the room, and the maximum ambient temperature is 35°.

Power source: electricity, three-phase AC 380/220 volts.

1

, Motor selection

(1), Motor type selection: Y series three-phase asynchronous motor

(2) , Motor power selection:

①Total efficiency of the transmission device:

=0.98×0.99 ×0.96×0.99×0.96

②Input required by the working machine Power:

Because F=0.2 KN=0.2 KN= 1908N

=FV/1000η

=1908×2/1000×0.96

< p>=3.975KW

③The output power of the motor:

=3.975/0.87=4.488KW

Let the rated power of the motor P = (1～1.3 )P, the rated power of the motor P = 5.5KW can be obtained by looking up the table.

⑶. Determine the motor speed:

Calculate the drum working speed:

= (60×v)/(2π×D/2)

= (60×2)/(2π×0.2)

=96r/min

Based on the recommended reasonable range of transmission ratio, take the cylindrical gear transmission first-stage reducer transmission Ratio range I' =3~6. Taking the V-belt transmission ratio I’ =2~4, the timing range of the total transmission ratio is I’ =6~24. Therefore, the optional range of motor speed is n' = (6~24)×96=576~2304r/min

⑷. Determine the motor model

According to the above calculation, it is within this range The synchronous speed of the motor is 1000r/min and 1500r/min. Taking into account the conditions of the motor and transmission device, and also reducing the weight and cost of the motor, the synchronous speed can be determined to be 1500r/min. According to the required rated power and synchronization The model of the motor determined by the speed is Y132S-4, and the full load speed is 1440r/min.

Its main performance: rated power: 5.5KW, full load speed 1440r/min, rated torque 2.2, mass 68kg.

2. Calculate the total transmission ratio and allocate transmission ratios at each level

(1). Total transmission ratio: i =1440/96=15

( 2) Allocate transmission ratios at each stage:

According to the instruction book, take gear i =5 (i=3~6 is reasonable for single-stage reducer)

=15/5=3

3. Calculation of motion parameters and dynamic parameters

⑴. Calculate the speed of each axis (r/min)

=960r/min

=1440/3=480(r/min)

=480/5=96(r/min)

⑵ Calculate the power of each axis (KW)

The rated power of the motor Pm=5.5KW

So

P =5.5×0.98×0.99=4.354KW

=4.354×0.99×0.96 = 4.138KW

=4.138×0.99×0.99=4.056KW

⑶ Calculate the torque of each axis (N?mm)

TI=9550×PI/nI= 9550×4.354/480=86.63N?m

=9550×4.138/96 =411.645N?m

=9550×4.056/96 =403.486N?m

3. Design calculation of transmission parts

(1) Design calculation of gear transmission

(1) Select gear material and accuracy level

Consider deceleration The transmission power of the gear is not large, so the gear adopts soft tooth surface. The pinion is quenched and tempered with 40Cr, and the tooth surface hardness is 240~260HBS. The large gear is made of 45# steel, quenched and tempered, and the tooth surface hardness is 220HBS; select level 7 precision according to the instruction book. Tooth surface roughness R ≤1.6~3.2μm

(2) Determine the relevant parameters and coefficients as follows:

Transmission ratio i

Take the number of pinion teeth Z =20. Then the number of teeth of the large gear:

=5×20=100, so take Z

Actual transmission ratio

i =101/20=5.05

Transmission ratio error: (i -i)/I= (5.05-5)/5=1%<2.5% Available

Gear ratio: u=i

Take Module: m=3; addendum height coefficient h =1; radial clearance coefficient c =0.25; pressure angle =20°;

Then h *m=3, h ) m=3.75

p>

h=(2 h)m=6.75, c= c

Dimension circle diameter: d =×20mm=60mm

d =3×101mm=303mm

Get φ from the instruction book

Tooth width: b=φ =0.9×60mm=54mm

=60mm,

b < /p>

Tooth tip circle diameter: d )=66,

d

Tooth root circle diameter: d )=52.5,

d ) =295.5

Base circle diameter:

d cos =56.38,

d cos =284.73

(3) Calculate the gear transmission Center moment a:

a=m/2(Z)=3/2(2101)=181.5mm Hydraulic winch≈182mm

(2) Design calculation of shaft< /p>

1. Design calculation of input shaft

⑴. Preliminary calculation of shaft diameter according to torque

Select 45# tempered, hardness 217~255HBS

< p>According to the instruction book and looking up the table, take c=110

So d≥110 (4.354/480) 1/3mm=22.941mm

d=22.941×(1+5 %)mm=24.08mm

∴Select d=25mm

⑵. Structural design of the shaft

①Positioning, fixing and assembly of parts on the shaft

p>

In a single-stage reducer, the gear can be arranged in the center of the box, symmetrically distributed relative to the two bearings. The left side of the gear is positioned by the shoulder, and the right side is axially fixed with a sleeve. The connection is fixed with a flat key as a transition fit. The bearings are positioned by the shaft shoulder and the large cylinder respectively, and are fixed by transition fit

②Determine the diameter and length of each section of the shaft

Section I: d =25mm

, L = (1.5～3)d, so the length is L

∵h=2c

c=1.5mm

+2h=25+2×2 ×1.5=31mm

Considering the gear end face and the inner wall of the box, there should be a certain distance between the end face of the bearing and the inner wall of the box.

Take the length of the sleeve to 20mm. The length of the shaft section passing through the sealing cover should be determined according to the width of the sealing cover and considering that there should be a certain moment distance between the coupling and the outer wall of the box. For this reason, take the length of this section to 55mm and install the gear section. The length should be 2mm smaller than the wheel hub width, so the length of section II is:

L = (2+255)=77mm

The diameter of section III:

Initial Select type 30207 angular contact ball bearing, whose inner diameter d is 35mm, outer diameter D is 72mm, and width T is 18.25mm.

=d=35mm, L =T=18.25mm, take L

Section IV diameter:

According to the manual: c=1.5

h=2c=2×1.5=3mm

The rolling bearing on the left side of this section Considering the positioning shoulder, it should be convenient for the disassembly of the bearing. The installation dimension h=3 obtained from the manual should be checked according to the standard. The diameter of this section should be: d = (35+3×2)=41mm

Therefore, section IV is designed into a stepped shape, with the diameter of the left section being 41mm

+2h=35+2×3=41mm

The length is the same as the sleeve on the right, which is L

p>

Diameter of section V: d =50mm., length L =60mm

Take L

The shaft support span L=80mm can be calculated from the length of each section of the shaft above < /p>

Diameter of section VI: d =41mm, L

Diameter of section VII: d =35mm, L

2. Design calculation of output shaft

⑴. Initial calculation of shaft diameter based on torque

Select 45# quenched and tempered steel, hardness (217~255HBS)

According to the formula on page P235 of the textbook (10- 2), Table (10-2) takes c=110

=110× (2.168/76.4) =38.57mm

Consider the keyway and increase the diameter by 5%, then

d=38.57×(1+5%)mm=40.4985mm

∴ Take d=42mm

⑵, Structural design of the shaft

① Positioning, fixing and assembly of shaft parts

In a single-stage reducer, the gears can be arranged in the center of the box and symmetrically distributed relative to the two bearings. The left side of the gear is positioned with a shoulder and the right side is positioned with a sleeve. The cylinder is positioned axially, and the circumferential positioning uses keys and transition fits. The two bearings are positioned by bearing shoulders and sleeves respectively. The circumferential positioning uses transition fits or interference fits. The shaft is stepped, and the left bearing is installed from the left. The gear The sleeve, right bearing and pulley are installed in sequence from the right.

② Determine the diameter and length of each section of the shaft

The primary 30211 angular ball bearing has an inner diameter d of 55mm, an outer diameter D=100mm, and a width T of 22.755mm. Considering the gear end face and the inner wall of the box, there should be a certain moment separation between the end face of the bearing and the inner wall of the box. If the sleeve length is 20mm, then the length of this section is 42.755mm. The length of the installed gear section is 2mm in width of the hub.

Then d =42mm L = 50mm

L = 55mm

L = 60mm

L = 68mm

L =55mm

L

IV. Selection of rolling bearings

1. Calculate the input bearings

Select type 30207 angular contact ball bearings , its inner diameter d is 35mm, its outer diameter D is 72mm, and its width T is 18.25mm.

2. Calculate the output bearing

Select the 30211 type angular ball bearing, its inner diameter d is 55mm, outer diameter D=100mm, width T is 22.755mm

5. Selection of key connection

1. The output shaft and pulley are connected by flat key connection

< p>Key type and size selection:

Pulley transmission requires good alignment between the pulley and the shaft, so C-type flat key connection is selected.

According to the shaft diameter d =42mm, L =65mm

Check the manual and choose the C-type flat key to get: Winch

Part No. 22 in the assembly diagram Select the GB1096-79 series key 12×56

The following results show: key width b=12, key height h=8, because the axis length L = 65, so the key length L=56

2. The output shaft and the gear are connected with a flat key

=60mm,L

Check the manual and choose a C-type flat key, you get:

< p>In the assembly drawing, the H?gglunds No. 36 part uses the GB1096-79 series key 18×45

Then it is found that: key width b=18, key height h=11, because the axis length L =53 , so the key length L=45

3. The input shaft and the pulley are connected using a flat key = 25mm L

Check the manual

Select Type A flat key, we get:

Part No. 29 in the assembly diagram uses the GB1096-79 series key 8×50

Then we find: key width b=8, key height h=7, Since the shaft length L = 62, the key length L = 50

4. The output shaft and the gear are connected with a flat key

=50mm

L < /p>

Check the manual

Select the A-type flat key and get:

Part No. 26 in the assembly diagram uses the GB1096-79 series key 14×49

Then it is found that: key width b=14, key height h=9, because the axis length L =60, so the key length L=49

6. Calculation of the main dimensions of the box and cover

The cabinet adopts a horizontally split structure and is cast from HT200 gray cast iron.

The main dimensions of the box are calculated as follows:

7. Bearing end cover

Main dimension calculations

Bearing end cover: HT150 d3=8

n=6 b=10

8. Reducer

Design of reducer accessories

1

, retaining ring : GB886-86

It was found that: inner diameter d=55, outer diameter D=65, retaining ring thickness H=5, right shoulder shaft diameter D1≥58

2

, Oil mark: M12: d =6, h=28, a=10, b=6, c=4, D=20, D

3

, Angle screw plug

M18

×

1.5: JB/ZQ4450-86

9.

Catalog of Design Reference Materials

1. Editors-in-Chief: Wu Zongze and Luo Shengguo. Mechanical Design Curriculum Design Handbook. Beijing: Higher Education Press, 1999.6

2. Editors: Xie Lanchang et al. Design of Precision Instrument Mechanisms .Hangzhou: Zhejiang University Press, 1997.11