Content

1.Taskforthecourseproject 1

2. KINEMATIC AND FORCE ANALYSIS OF A MECHANICAL DRIVE 2

For our case 2

3. Analysis of allowable stresses 5

5

4. ANALYSIS OF THE STRAIGHT SPUR GEARS FOR STRENGTH 8

5. ANALYSIS OF THE CHAIN DRIVE 11

6. Analysis of shafts 15

Standard lip seals (GOST 8752-79) 16

7. Shaft analysis for strength 18

Values of [S] 23

8. Analysis of the rolling contact bearings 24

9.CALCULATION OF KEYED JOINTS. 26

Standard Sunk Keys 26

10. Designing the gear speed reducer 28

Airvent 29

Dip sticks 30

11. DESIGNING THE MECHANICAL DRIVE 31

Standard УM series bearing housings (GOST 13218.3-80), mm 31

12. REFERENCE LIST 33

13. Appendixes 34

1.Taskforthecourseproject Task № 2

Design a mechanical drive of a chain conveyer for processing single and small size cargo in airports.

M echanical Drive Diagram:

Initial Data:

Service life – 8 years; К_daily = 0.25; К_annual = 0.4; constant load. F_t=3 kN; Peripheral speed V=1.9 m/sec; diameter of the working organ Sprocket D=300 mm.

2. Kinematic and force analysis of a mechanical drive

2. 1. Determine the output power of the drive

P_out = F_t∙V= 3∙1.9=5.7 kW

2. Determine the total efficiency of the drive

For our case

 = _c_ssg_cd_b³,

where_cis the efficiency of the coupling;

_ssg is the efficiency of straight spur gears;

_cd is the efficiency of the chain drive;

_b takes into account losses in one pair of bearings.

The following components are used for mechanical drive being analyzed:

• belt drive with flat belt;

• chain drive with roller chain;

• rolling bearings;

• coupling with rubber bushed studs;

Let us assume that _c = 0.996, _ssg= 0.99, _cd = 0.96,_b= 0.995. Then

= 0.996 0.99 0.96 0.995³ = 0.932.

3. Determine the input power.

.

2.4. Select the electrical motor.

For our mechanical drive we select 4A132M6 Induction Motor (P_r= 7.5 kW, n_s= 1000 rpm, S = 3.2 %.).

2.5. Determine the motor rated rotational speed n_r

2.6. Determine the output rotational speed

2.7. Determine the total velocity ratio of the mechanical drive

2.8. Distribute the total velocity ratio between mechanical drive steps.

The total velocity ratio can be found by the formula

u = u_redu_cd ,

whereu_redis the speed reducer velocity ratio; u_cd is the chain drive velocity ratio.

Let us take u_red=u_ssg= 3.55.

Determine the velocity ratio of the open drive. In our case it is chain drive

2.9.Determine the rotational speed of all shafts

n₁ = n_r= 968 rpm;

Obtained magnitude of n₃ must be equal to n_out calculated according to p. 2.6. Error ε must be not more than 1%. In our case ε=0.007%.

2.10.Determine the angular velocity of all mechanical drive shafts:

2.11.Determine the power at mechanical drive shafts.

P₁= P_inp_c_b= 6115.880.9960.995 = 6.061 kW;

P₂= P₁_ssg _b= 6.0610.990.995 = 5.970 kW;

P₃= P₂_cd _b= 5.9700.960.995 = 5.703 kW;

Obtained magnitude of P₃ must be equal to P_out calculated according to the p.2.1. Error should not be greater than 1%.

In our case ε=0.05%.

2.12.Determine the torques at all shafts.

Checking.

The output torque T_outcan also be found as

Determine the error. It should be less than 1%.

3. Analysis of allowable stresses

Let us analyze the speed reducer of the mechanical drive when:

n^p = 968 rpm;n^g= 272.676 rpm.

3.1. Select the material of toothed wheels.

A pinion and a gear are produced from identical carbon or alloy steel, such as Steel 45 (0.45C), Steel 40X (0.40C-Cr), Steel 40XH (0.40C-Cr-Ni). Heat treatment of both the gear and the pinion is martempering. The pinion hardness is ranged from 269 to 302 BHN and the gear hardness is ranged from 235 to 262 BHN.

3.2. Determine the mean magnitude of the hardness of the gear and the pinion:

- for the pinion

- for the gear

In our case a pinion and a gear are produced from identical alloy steel of grade Steel 40XH.

BHN; BHN.

3.3. Determine the allowable contact stress for the pinion and for the gear.

1. 3.3.1.Determine the limit of contact endurance for the pinion and for the gear .

3.3.2.Determine the base number of stress cycles for the pinion and for the gear .

2. stress cycles;

stress cycles.

3.3.3. Determine the service life in hours for the gearing:

t = L · 365 · K_a · 24 · K_d,

whereL is the service life in years; K_a is the annual utilization factor which is used to take into account use of the gearing during a year; K_d is the daily utilization factor which is used to take into account use of the gearing for 24 hours.

In our case:

the service life of the gearing is 8 years, K_a= 0.25, K_d =0.4

.

3.3.4. Determine the design number of stress cycles for the pinion and the gear.

wheren^p and n^g are correspondingly rotational speeds of the pinion and the gear; c is the number of gears meshing with the gear being analysed.

In our case c = 1, n^p = 968 rpm; n^g = 272.676 rpm.

.

3.3.5. Determine the durability factor for the pinion and for the gear.

if N_Hi ≥ N_HO then K_HL=1,

ifN_Hi< N_HO then .

In our case

, , ,

consequently

, , ,

consequently

3.3.6. Determine the safety factor S_H for the pinion and for the gear.

- for homogeneous structure of the material

(heat treatment is normalizing, martempering

and full hardening) S_H = 1.1;

3.3.7. Determine the contact allowable stresses for the gear and for the pinion

.

In our case:

.

If H^p-H^g 70BHN, we assume as the design allowable contact stress the less magnitude of above calculated stresses, where H^p and H^g are correspondingly hardness of the pinion and gear materials.

Otherwise, the design allowable contact stress is determined by the following formula

.

In our case H^p-H^g< 70BHN. That is why for further calculations we take as the design allowable contact stress .

4. Determine the allowable bending stresses of the pinion and for the gear.

   1. Determine the limits of the bending endurance for the pinion and for the gear .

.

3.4.2. Determine the base number of stress cycles .

For steels = 4∙10⁶.

3.4.3. Determine the design number of stress cycles for the pinion and the gear.

;

.

3.4.4. Determine the durability factor for the pinion and for the gear.

if N_bi ≥ N_b0 then K_bL=1,

ifN_bi< N_b0 then ,

where m=3 for toothed wheels with hardness H ≤ 350 BHN and m=9 if H > 350 BHN.

In our case: , consequently

, consequently

3.4.5. Determine the safety factor S_b for the pinion and for the gear.

- for wheels made of forged blanks (our case) S_b = 1.75;

3.4.6. Determine the bending allowable stresses for the gear and for the pinion

, .

In our case:

For further calculations, we assume as the design allowable bending stress the less magnitude of above calculated stresses .