design project of transmission system case study 10 plan 1

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HO CHI MINH UNIVERSITY OF TECHNOLOGYVIETNAM NATIONAL UNIVERSITY

FALCUTY OF MECHANICALENGINEERING DEPARTMENT OF

HO CHI MINH CITY, October 4 2021th

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Mechanical Faculty, Ho Chi Minh city University of Technology

Nowadays, the development of technology has received growing interest bygovernments from many countries. With the help of technology, we can easily increaseour productivity and create more safe conditions in workplaces. To reach theseaccomplishments, it takes us a lot of time to practice, do research and improve ourtechnological skills from the beginning of our university life.

Design project of transmission system is a helpful subject which allows mechanicalengineering students to apply some theoretical knowledge from the previous subjectmachine elements. It also helps students to get familiar with calculating each elementfor a realistic mechanical design.

Thanks to Dr Nguyen Tan Tien, with his instruction, I could totally complete thisproject. During the period of doing this project, I could reinforce all knowledge,approach and learn how to use some engineering softwares such as Autocad,Solidworks, etc. Also, I was able to learn to accomplish the mindset of an engineerwhich plays a pivotal role in my future career.

I have tried my best to perform this project. However, due to limited time and mylack of practical experience, there still are some minor errors in this report. Therefore, Ilook forward to receiving suggestions in order to enrich my knowledge.

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4. Technical specification table...4

CALCULATION AND DESIGN OF MACHINE ELEMENTS...5

I. Calculation and design of -belt transmission system...5𝑉1. Select belt and belt diameter 2...5

2. Define preliminary center distance...5

3. Define belt length and center distance...6

4. Calculating the number of belts...7

5. Define main dimension of pulley...7

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Mechanical Faculty, Ho Chi Minh city University of Technology

3. Define pitch diameter...12

4. Analyze contact stress...14

5. Analyze bending stress...15

6. Analyze overload stress...16

7. Force calculating...17

8. Summary...17

III. Design elastic coupling...18

1. Choose dimension...18

2. Force apply on the coupling...18

ANALYZING LUBRICATION CONDITION OF THE SYSTEM...19

I. Lubrication condition of the system:...19

II. Summary...19

CALCULATION AND DESIGN OF SHAFT AND KEY...20

I. Choose shaft material:...20

II. Calculate and design shaft 𝐼...21

1. Calculate forces distribute on shaft 𝐼...21

2. Force and momentum analysis on shaft 𝐼...22

3. Sections diameter and keys selection:...24

III. Calculate and design shaft 𝐼𝐼...25

1. Calculate forces distribute on shaft 𝐼𝐼...25

2. Force and momentum analysis on shaft 𝐼𝐼...25

3. Sections diameter and keys selection:...28

IV. Shafts and keys test...29

1. Test for shafts...29

2. Test for keys...31

CALCULATION AND TEST BEARINGs...32

I. Calculation and select bearing for shaft 𝐼...32

1. Calculation dynamic loading...32

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2. Calculation static loading...33

II. Calculation and select bearing for shaft 𝐼𝐼...33

1. Calculation dynamic loading...34

2. Calculation static loading...35

DESIGN GEAR BOX COVER AND RELATIVE ELEMENT SELECTION... 36

I. Calculate gear box dimensions...36

II. Calculation relative elements...37

6. Oil level stick...40

7. Oil seal and oil barrier...40

Tolerance and kind of assembly selection...42

I. Tolerance of key and keyway...42

II. Tolerance assembly of the gears...42

III. Tolerance assembly of bearings...42

IV. Tolerance assembly of oil seal...43

V. Tolerance assembly of caps of hub...43

REFERENCES...44

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INDEX OF IMAGES AND DIAGRAMS

Figure 1. 1: Position of each shaft diagram 2

Figure 2. 1: Classical 𝑉-belt...5

Figure 4. 1: Shafts rotational direction diagram...20

Figure 4. 2: Force equilibrium diagram on 2 shafts...21

Figure 4. 3: Force distribution diagram on shaft 𝐼...22

Figure 4. 4: Free body diagram and shaft structure of shaft 𝐼...23

Figure 4. 5: Drawing of shaft 𝐼...25

Figure 4. 6: Force distribution diagram on shaft 𝐼𝐼...26

Figure 4. 7: Free body diagram and shaft structure of shaft 𝐼𝐼...27

Figure 4. 8: Drawing of shaft 𝐼𝐼...29

Figure 6. 1: Eyebolt...37

Figure 6. 2: Taper pin...38

Figure 6. 3: Oil plug...38

Figure 6. 4: Visitor door...39

Figure 6. 5: Vent button...39

Figure 6. 6: Oil level stick...40

Figure 6. 7: Oil barrier...40

Figure 6. 8: Oil seal...41

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INDEX OF TABLES

Table 1. 1: Designing...1

Table 1. 2: Efficiency of each transmission parts...1

Table 1. 3: Motor information...2

Table 1. 4: Power calculating...3

Table 1. 5: Rated speed calculating...3

Table 1. 6: Torque calculating...3

Table 1. 7: The overall parameters of the shafts...4

Table 2. 1: Input data of 𝑉-belt system...5

Table 2. 2: The overall parameters of -belt type 𝑉 𝐴...5

Table 2. 3: The parameters of -belt pulley type 𝑉 𝐴...7

Table 2. 4: Summary specification of the 𝑉-belt transmission...9

Table 2. 5: Input data of bevel gear system...10

Table 2. 6: Summary specification of the bevel gear transmission...17

Table 2. 7: Input data of elastic coupling...18

Table 2. 8: Dimensions of elastic coupling...18

Table 2. 9: Dimensions of the pin...18

Table 4. 1: Summary specification of the oil lubrication...19

Table 4. 2: Shaft material mechanical properties...20

Table 4. 3: Force magnitude distribute on shaft 𝐼...22

Table 4. 4: Key statistic on shaft 𝐼...24

Table 4. 5: Force magnitude distribute on shaft 𝐼𝐼...25

Table 4. 6: Key statistic on shaft 𝐼𝐼...28

Table 4. 7: Shaft test at dangerous sections...30

Table 4. 8: Key statistic on both shafts...31

Table 4. 9: Key test for all cross sections...31

Table 5. 1: Forces apply on shaft 𝐼...32

Table 5. 2: Bearing size preliminarily selection for shaft 𝐼...32

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Mechanical Faculty, Ho Chi Minh city University of Technology

Table 5. 3: Forces apply on shaft 𝐼𝐼...34

Table 5. 4: Bearing size preliminarily selection for shaft 𝐼𝐼...34

Table 6. 1: Eyebolt dimension...37

Table 6. 2: Lock pin dimension...38

Table 6. 3: Oil plug dimension...38

Table 6. 4: Visitor door dimension...39

Table 6. 5: Vent button dimension...40

Table 6. 6: Oil seal dimension...41

Table 7. 1: Tolerance of key and keyway...42

Table 7. 2: Tolerance assembly of the gears...42

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1. Efficiency of whole system

Based on table 2.3 [1], we preliminarily select the efficiency of each transmission parts.

Table 1. 2: Efficiency of each transmission parts

Transmission parts Efficiency factor

Flexible coupling, 𝜂𝑐 0.98Roller bearing, 𝜂𝑟𝑏 0.99

Therefore, we can determine the efficiency of transmission system based on equation2.9 [1]:

3. Distribute transmission ratio

Equation 2.15 [1] is used to determine the total ratio:𝑢𝑡 = 𝑢𝑏 × 𝑢𝑏𝑔 = 4 × 2.5 = 10.Where: 𝑢𝑏 = 4 velocity ratio of -belt; (table 2.4 [1]).𝑉

𝑢𝑏𝑔 = 2.5 velocity ratio of bevel gearbox.

From equation 2.18 [1], the number of preliminary revolutions can be determined as follow:

𝑛𝑝 = 𝑛𝑙𝑣 × 𝑢𝑡 = 160 × 10 = 1600 𝑟𝑝𝑚.

We choose number of synchronous revolutions of motor 𝑛𝑠 = 1500 𝑟𝑝𝑚.

4. Select motor

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Motor Shaft

Shaft 𝐼Shaft 𝐼𝐼

Working Shaft

Mechanical Faculty, Ho Chi Minh city University of Technology

2We choose a motor that satisfy the conditions 2.19 [1]:

{𝑃𝑁 ≥ 𝑃𝑛𝑛𝑁 ≅ 𝑛𝑠

We choose motor 𝑆𝑔 100 − 4𝐿 𝐵, followed by [2]. The motor has rated output power𝑃𝑁 = 3 𝑘𝑊 and rated speed 𝑛𝑁 = 1415 𝑟𝑝𝑚.

Table 1. 3: Motor information

Type 𝑃 𝑘𝑊, 𝑛, 𝑟𝑝𝑚 𝑇𝐿𝑇𝑁

𝑇𝑁 𝐽, 𝑘𝑔𝑚2 𝑚 𝑘𝑔,𝑆𝑔 100𝐿 − 4𝐵 3 1415 2.6 2.7 0.0065 26

After selecting the motor, we recalculate the transmission ratio:⎛ 𝑢𝑏𝑔 = 2.5

𝑢𝑡⎨ =𝑛𝑁

𝑛𝑙𝑣1415= 8.84 ; 160= { 𝑢𝑡 = 𝑢𝑏 × 𝑢𝑏𝑔

𝑢𝑡⇒ 𝑢𝑏 =

𝑏𝑔8.84= = 3.54.

It is satisfy the range of transmission ratio for V-belt from table 2.4 [1].

II. Technical specification

Figure 1. 1: Position of each shaft diagram

𝑢

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After we preliminary determine the efficiency of each transmission parts, power of themotor and ratio transmission of system, we have enough data to figure out power, ratedspeed of each shaft and to create a Technical Specification Table.

1. Power

Table 1. 4: Power calculating

Power transmitted on

𝑘𝑊Power transmitted on

Shaft 𝐼 𝑃𝐼 𝑃𝐼 = 𝑃𝑁 × 𝜂𝑏 × 𝜂𝑟𝑏𝑃𝐼 = 3 × 0.96 × 0.99 = 2.85Power transmitted on

𝑃𝐼𝐼 = 𝑃𝐼 × 𝜂𝑏𝑔 × 𝜂𝑟𝑏𝑃𝐼𝐼 = 2.85 × 0.96 × 0.99 = 2.71Power transmitted on

Working Shaft 𝑃𝑊𝑆.

𝑃𝑊𝑆 . = 𝑃𝐼𝐼 × 𝜂𝑐 × 𝜂𝑟𝑏𝑃𝑊𝑆 . = 2.85 × 0.98 × 0.99 = 2.6

2. Rated speed

Table 1. 5: Rated speed calculating

Rated Speed on Motor

𝑟𝑝𝑚Rated speed on Shaft 𝐼

𝑛𝑁 1415𝑛𝐼 =

𝑢 = = 4003.54 𝑏Rated speed on Shaft 𝐼𝐼

𝑛𝐼 400𝑛𝐼𝐼 =

𝑢 = = 1602.5 𝑏𝑔Rated speed on Working

𝑛𝐼𝐼 160𝑛𝑊𝑆 . =

𝑢 =1 = 160𝑐

3. Torque

Table 1. 6: Torque calculating

Torque transmittedon Motor Shaft 𝑇

𝑇 = 9.55 × 10 × 6𝑃𝑁

𝑁3𝑇𝑁 = 9.55 × 10 ×6 = 20247.35

𝑁𝑚𝑚Torque transmitted

on Shaft I

𝑃𝐼𝑇𝐼 = 9.55 × 10 ×6

𝑛𝐼2.85𝑇𝐼 = 9.55 × 10 6 × = 68072.4400

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Mechanical Faculty, Ho Chi Minh city University of Technology

4Torque transmitted

on Shaft II

𝑃𝐼𝐼𝑇𝐼𝐼 = 9.55 × 10 ×6

𝑛𝐼𝐼2.71𝑇𝐼𝐼 = 9.55 × 10 6 × = 161740.02 160Torque transmitted

on Working Shaft

𝑃𝑊𝑆.𝑇𝑊𝑆 . = 9.55 × 106 ×

𝑛𝑊𝑆.2.6𝑇𝐼𝐼𝐼 = 9.55 × 10 6 × = 156920.17 160

4. Technical specification table

Table 1. 7: The overall parameters of the shafts

Motor shaft Shaft 𝐼 Shaft 𝐼𝐼 WorkingShaft

Rated speed , 𝑛 𝑟𝑝𝑚 1415 400 160 160Torque , 𝑇 𝑁𝑚𝑚 20247.35 68072.4 161740.0

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CALCULATION AND DESIGN OF MACHINE ELEMENTS

I. Calculation and design of -belt transmission system𝑽

1. Select belt and belt diameter 2

From the table 1.8, we summary the input data of the belt transmission.

Table 2. 1: Input data of -belt system𝑽

Rated speed , 𝑛 𝑟𝑝𝑚 1415Torque , 𝑇 𝑁𝑚𝑚 20247.35

Based on figure 4.1 and table 4.13 [1], we choose -belt type 𝑉 𝐴 with the following parameters:

Figure 2. 1: Classical -belt 𝑽

Table 2. 2: The overall parameters of -belt type 𝑽 𝑨

Type belt Cross-sectional dimensions, 𝑚𝑚 Sectional area𝐴, 𝑚𝑚2 Weight𝑞𝑚, /𝑘𝑔𝑚

2. Define preliminary center distance

Based on the standard diameter of pulley, we choose 𝑑1 = 112 𝑚𝑚. Belt velocity:

𝑣 = 𝜋𝑑 𝑛1 1 =60000

𝜋 × 112 ×141560000

𝐻 𝑦0

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Mechanical Faculty, Ho Chi Minh city University of Technology

30𝑙13 = 75 + 5 + 5 + 35 + 0.5 × 17 − 38 × cos 21.8° = ( )

110.86 𝑚𝑚.

With: 𝑘1 = 5 𝑚𝑚follow table 10.3 [1];𝑘2 = 5 𝑚𝑚 follow table 10.3 [1];

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𝑌𝑂𝑋 𝑍𝐹

𝑙𝑚13 = (1.2 ÷ 1.4)𝑑𝐼 = (1.2 ữ 1.4) ì 28 = (33.6 ữ 39.2) 𝑚𝑚 by formula10.10 [1]

We choose 𝑙𝑚13= 35 𝑚𝑚. Then we choose 𝑙13 = 111 𝑚𝑚.

2. Force and momentum analysis on shaft :𝑰

Table 4. 3: Force magnitude distribute on shaft 𝑰

Magnitude 631.23 1867.88 252.49 634.54Therefore, we get the force distribution diagram as below:

Figure 4. 3: Force distribution diagram on shaft 𝑰Reaction force at pins:

As figure 3.3, we have total force and momentum equations on shaft , yOz plane:𝐼∑ 𝐹𝑦 = 𝐹𝑟 1− 𝐹𝐵𝑦 − 𝐹𝐶𝑦 + 𝐹𝑟 2= 0

∑𝑀𝑥⁄𝐵 = 𝐹𝑟1𝑙12 + 𝐹𝐶𝑦𝑙11 +𝐹𝑎2𝐹𝐵𝑦 = 885.72 𝑁

2 − 𝐹𝑟2𝑙13 = 0⇔ {

𝐶𝑦= 380.04 𝑁

Where: 𝑑𝑚 1= 72.89 𝑚𝑚 is the mean cone diameter of bevel gear.

As figure 3.3, we have total force and momentum equations on shaft , xOz plane:𝐼{ ∑ 𝐹𝑥 = 𝐹𝐵𝑥− 𝐹𝐶𝑥+ 𝐹𝑡 2=

∑𝑀𝑦⁄𝐵 𝐹 = 𝐶𝑥 𝑙11 − 𝐹𝑡2𝑙13 =0

⇔ { 𝐹𝐵𝑥 = 896.58 𝑁𝐹𝐶𝑥 = 2764.46 𝑁

Therefore, we get momentum on yOz, xOz plane and torque diagram on shaft as 𝐼 below:

→𝐶𝑦𝐹→𝑡2𝑙12 𝑙11

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𝐶𝑥𝐹→𝑡 2

𝐹→𝑟 2

𝐹→𝐵 𝑥𝐹→𝐵𝑦

𝐹→𝐶 𝑦 →𝑚 1

× 𝐹→2 𝑎2

Figure 4. 4: Free body diagram and shaft structure of shaft 𝑰𝑌

𝑂 𝑍𝑋

𝑂 𝑍𝑋

+−+−

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Total bending momentum on shaft calculated by formula 10.15 [1]:𝐼 𝑀𝐶 = √ 2𝑀 + 2𝑀 = 13522.582 + 67243.682 = 68589.89√

𝑀𝐷 = 9202 𝑁𝑚𝑚.

Equivalent momentum on shaft calculated by formula 10.16 [1]:𝐼 𝑀𝑒𝐴 = √0.75𝑇2 = 0.75 × 68072.4 = 58952.43 √ 2 𝑁𝑚𝑚.𝑀𝑒𝐶 = √𝑀2 + 0.75𝑇2 = √68589.892 + 0.75 × 68072.42

3. Sections diameter and keys selection:

Shaft diameter at sections calculate by formula 10.17 [1]:𝐼

With C45 steel, preliminary diameter 𝑑𝐼 = 28 𝑚𝑚, we have allowable stress [𝜎] = 63 𝑀𝑃𝑎:

3 𝑀𝑒𝐴 3 58952.43𝑑𝐴 = √

0.1 × [𝜎] = √ 0.1 ×63

= 21.07 𝑚𝑚.

3 𝑀𝑒𝐶 3 90443.14𝑑𝐶 = √

0.1 × [𝜎] = √ 0.1 ×63

= 24.3 𝑚𝑚.

3 𝑀𝑒𝐷3 59666.28𝑑𝐷 = √

0.1 × [𝜎] = √ 0.1 ×63

= 21.16 𝑚𝑚.

There are keyseats at section and . Therefore:𝐴 𝐷𝑑𝐴 ≥ 1 + 4% × 21.07 = 21.91 ( ) 𝑚𝑚.𝑑𝐷 ≥ 1 + 4% × 21.16 = 22.01 ( ) 𝑚𝑚.

However, in reality, we must consider the technical of assembling and the toleranceof the shaft, we choose the diameter of shaft segments according to standard:𝐼

𝑑𝐴 = 22 𝑚𝑚 𝑑, 𝐶 = 35 𝑚𝑚 𝑑, 𝐷 = 25 𝑚𝑚.

Then we choose keys size for shaft which following table 9.1a [1]:𝐼

Table 4. 4: Key statistic on shaft 𝑰

Key dimension Keyway depth

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34Shaft diameter,

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Figure 4. 5: Drawing of shaft 𝑰

III. Calculate and design shaft 𝑰𝑰:

1. Calculate forces distribute on shaft 𝑰𝑰

Shaft preliminary diameter can be calculated by formula 10.4 [1]:𝐼𝐼 3 5𝑇𝐼𝐼 3 5 × 161740.02

𝑑𝐼𝐼 = √ [ ]𝜏 = √

= 31.86 𝑚𝑚.

Where: stresses package [𝜏] = 25 𝑀𝑃𝑎.

We choose preliminary diameter for shaft 𝐼𝐼is 𝑑𝐼𝐼 = 32 𝑚𝑚. Based on the preliminary diameter we get that 𝑏0 = 19.

Determined the distance between bearing supports and force placements:From table 10.4 [1], we calculate:

Distance from bevel gear to bearing 0 on shaft 𝐼𝐼:𝑙22 = 0.5(𝑏0 + 𝑏 𝑐𝑜𝑠13 𝛿2) + 𝑘1 + 𝑘2

𝑙22 = 0.5 × 19 + 33 × ( 𝑐𝑜𝑠 68.2° + 10 + 10 = 35.5 ) 𝑚𝑚.With: 𝑘1 = 10 𝑚𝑚 follow table 10.3 [1];

𝑘2 = 10 𝑚𝑚 follow table 10.3 [1];

Distance between 2 bearings should be determined so that the bearings symmetrize atthe center of bevel pinion:

𝑙21 = 2𝑙22 + 𝑑𝑚 1= 2 × 35.5 + 72.89 = 143.89 𝑚𝑚.Then we choose distance between 2 bearings 𝑙21 = 144 𝑚𝑚; Distance between bearing 0 to haft of the elastic coupling on shaft II:

𝑙23 = 0.5(𝑏0 + 𝑙m23) + 𝑙21 + 𝑘3 + ℎn

𝑙23 = 0.5 × (19 + 45) + 144 + 10 + 15 = 201 𝑚𝑚.With: ℎ𝑛 = 15 𝑚𝑚based on 10.3 [1];

𝑘3 = 10 𝑚𝑚 based on 10.3 [1].

2. Force and momentum analysis on shaft : 𝑰𝑰

Table 4. 5: Force magnitude distribute on shaft 𝑰𝑰

Force 𝐹 Gear (Bevel gear) Flexible coupling3

𝑟, 𝑁 𝐹𝑡3, 𝑁 𝐹𝑎3, 𝑁 𝐹𝑡4, 𝑁 𝐹𝑟4, 𝑁Magnitude 252.49 1867.88 631.23 3234.8 808.7

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Mechanical Faculty, Ho Chi Minh city University of Technology

36Therefore, we get the force distribution diagram as below:

Figure 4. 6: Force distribution diagram on shaft 𝑰𝑰Reaction force at pins:

As figure 3.4, we have total force and momentum equations on shaft , yOz plane:𝐼𝐼∑ 𝐹𝑦 = 𝐹𝐴𝑦 − 𝐹𝑟 3− 𝐹𝐶𝑦 =

∑𝑀𝑥/𝐴 = 𝐹𝑟3𝑙22 −𝐹𝑎3

𝑑𝑚2

2 + 𝐹𝐶𝑦𝑙21 = 0

⇔ {

𝐶𝑦= 337.13 𝑁

Where: 𝑑𝑚 2= 182.22 𝑚𝑚 is the mean cone diameter of bevel gear.

As figure 3.4, we have total force and momentum equations on shaft , xOz plane:𝐼𝐼{ ∑ 𝐹𝑥 = 𝐹𝐴𝑥 − 𝐹𝑡 3− 𝐹𝐶𝑥 + 𝐹𝑟 4

= 0

∑𝑀𝐴 = 𝐹𝑡3𝑙22 + 𝐹𝐶𝑥 𝑙21 − 𝐹𝑟4𝑙23 = 0

⇔ {𝐹𝐴𝑥 = 1727.5 𝑁𝐹𝐶𝑥 = 668.33 𝑁

Therefore, we get momentum on yOz, xOz plane and torque diagram on shaft II as below:

𝐹

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2 𝑎3

𝐹→𝑟 336578.98 𝑁𝑚𝑚

𝑀𝑥20931.63

61326.47 𝑁𝑚𝑚161740.02 𝑁𝑚𝑚+

46095.9 𝑁𝑚𝑚𝑟4𝐹→𝐹→𝐶𝑦

𝑌𝑌𝑂 𝑍𝑋

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Mechanical Faculty, Ho Chi Minh city University of Technology

38From figure 3.7, we obtain the dangerous cross sections on shaft 𝐼𝐼:

Total bending momentum on shaft are calculated by formula 10.15 [1]:𝐼𝐼 𝑀𝐵 = √ 2𝑀 + 2𝑀 = 36578.982 + 61326.472 = √71406.99 𝑁𝑚𝑚.

3. Sections diameter and keys selection:

Shaft I diameter at sections calculate by formula 10.17 [1]:

With C45 steel, we get allowable stress [𝜎] = 63 𝑀𝑃𝑎from table 10.5 [1]. Therefore, the actual diameter can be determined, following equation 10.17 [1]:

3 𝑀𝑒𝐵 3 157222.24𝑑𝐵 = √

0.1 × [𝜎] =

= 29.22 𝑚𝑚.

3 𝑀𝑒𝐶 3 147460.87𝑑𝐶 = √

0.1 × [𝜎] =

= 26.61 𝑚𝑚.

3 𝑀𝑒𝐷3 140070.97𝑑𝐷 = √

However, in reality, we must consider the technical of assembling and the toleranceof the shaft, we choose the diameter of shaft segments according to standard:𝐼𝐼

𝑑𝐵 = 50 𝑚𝑚 𝑑, 𝐶 = 40 𝑚𝑚 𝑑, 𝐷 = 30 𝑚𝑚.

Then we choose keys size for shaft , based on the actual diameter and table 9.1a [1]:𝐼𝐼

Table 4. 6: Key statistic on shaft 𝑰𝑰

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