Report of design project topic 1 – plan 27

2.2 Calculating and reinforce calculation of Helical Gear System .... Chapter 2: Design and Calculate Spur Gear System .... 3.2 Calculating and reinforce calculation of Spur Gear .... 9.

VIETNAM NATIONAL UNIVERSITY HO CHI MINH CITY

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY

Class: CC01–ME3139 Instructor: LÊ THANH LONG

Nguyễn Tr n Minh Quân ầ 1952944

3) Two-stage cylindrical gear reducer

4) Flexible shaft coupling

5) Chain conveyer

• Rotates in one direction, Impact load is light, One shift is 8 hour

Tensile force F, N 7700

Velocity of the conveyer v, m/s 0.8

Diameter of the driving pulley D, mm 9

Working day per year Kng , day 220

ABSTRACT

In the day and age, technological breakthroughs have played an important role in the development of each country Applying technology to manufacture not only increases productivity effect but also creates more safety workplaces for humans To reach our achievements in the future, we need to spend more time practicing, researching, and improving our technical skills from the beginning of our university life

Machinery Elements Project is a subject which helps mechanical students familiar with designing and calculating kinematics and dynamics of a machine This is a challenge for students to use some engineering software such as Solid-works, AutoCAD, MATLAB, etc This subject also helps students apply knowledge from previous subjects like Theory

of Machine, Machine Element, etc

With the instruction, advice and teaching of the instructor Lê Thanh Long during the period of doing this project, I can reinforce all the knowledge, approach and be taught how to use engineering software

INDEX

ABSTRACT

Part 1: Selecting Motor and Distributing Ratio Transmission 1.1 Determine the engine

1.2 Determine the speed ratio

1.3 Characteristic table

Part 2 : Machine detail calculation Chapter 1: Design and Calculate V belt System

2.1 Input technical specification data

2.2 Calculating and reinforce calculation of Helical Gear System

2.3 Summary

Chapter 2: Design and Calculate Spur Gear System

3.1 Input technical specification data

3.2 Calculating and reinforce calculation of Spur Gear

3.3 Summary

Chapter 3: Design a V-belt transmission system

4.1 Input technical specification data

4.2 Material Selection and Type of V-belt

4.3 Calculating of V-belt system

4.4 Force calculation on V-belt

Chapter 4 Design and calculate Shaft in the system

5.1 Choose material for designing

5.2 Design and calculating Shaft I

5.3 Design and calculating Shaft II

5.4 Design and calculating shaft III

5.4.1 Determine the diameter and distance of each element on the shaft

5.4 Re-calculating Shaft and Key

Chapter 6: Design and calculate flexible coupling

Chapter 7: Design and calculate bearing in the system

7.1 Design and selecting the bearing of Shaft I

7.2 Design and selecting the bearing of Shat II

7.3 Design and selecting the bearing of shaft III

Chapter 8: Design a housing of gear box and some elements

8.1 Design a housing of a gearbox

8.2 Some engagement of some elements in the gear box

Chapter 9: Tolerance and kind of assembly selection

9.1 Tolerance of key and key way

9.2 Tolerance assembly of helical and spur gear

9.3 Tolerance assembly of bearing

9.4 Tolerance assembly of oil seal

9.5 Tolerance assembly of caps of hub

PART1 Selecting Motor and Distributing Ratio Transmission

1.1DETERMINE THE ENGINE

1.1.1 Calculating the necessary power :

 Transmission efficiency :

+ we choose :

o The efficiency of belt drive :  = 0.95

o The efficiency of gear drive :  = 0,98

o The efficiency of shaft coupling : shaft = 0.98

o The efficiency of bearing : bearing = 0.99

 =  * ( gear )2 * shaft* (bearing)5

= :#0∗ #53( );.< #53#83'80∗ '83(;.=)0∗ #8

 The angular velocity of the conveyer : KL@= 60000 M.2@ = 60000 (=∗##;;.B) ≈ 48.5 rpm

V : velocity of the conveyer

z : the number of gear teeth

t : Pitch of chain

 We choose the speed ratio:

- Speed ratio of the gear reducer: = 12 (6-30)

- Speed ratio of belt drive:  = 3 (3-5) The preliminary speed ratio of the system:

Usb = Ugr.Ubelt = 12.3 = 36

• The necessary angular velocity of the motor :

Ksb = KEKOP.u = 48,5 * 36 = 1746 rpm

1.1.3 Determine the motor :

♦ From the necessary power and the necessary angular velocity calculated, we choose

the standard electrical motor form the table P1.3 page 236 - Tính toán h ệthống d n ẫ

động c khí 1 (Trịnh Chất): ơ

- Condition : Pdc ≥ Pct (6,123 kW)

ndc ≈ nsb (1746 rpm )

Type Power (kW) Angular velocity

Checking the starting condition:

- The starting moment of the electrical motor must be larger than the initially drag moment of the load:

- YY. = 1 => Y[

Y \] ≥

1 Rℎ !^K EK_^^K ^! !!^`^_ E ! Y[

Y \]= 2 1.2.Distribute the speed ratio

- Speed ratio of the primary gear pair: = 4 1

- Speed ratio of the secondary gear pair: = 3 2

1.2.2 Calculate the angular velocities on shafts:

- Error between nIII and nlv : IB.5lIB.5

IB.5 * 100 = 0%

1.1.3 Distribute the powers on shafts:

• PIII = >mm∗ ηgear ∗ηbearing = 6.64(kW)

• PII = >m∗ ηgear ∗ηbearing = 6.84(kW)

• PI = >So2op∗ ηbelt ∗ ηbearing = 0.95 ∗ 0.99 ∗ 7.5 = 7.05

• Pchain= 6.381.2.4 Calculate torques on shafts:

PART 2 : MACHINE DETAIL CALCULATION:

1:Belt Drive design

đai nh dỏ 1 , mm

Chiều dài gi i ớhạn l, mm

bt b h y0

B 14 17 10.5 4.0 138 140-280 800-6300

- d1 = 1,2dmin = 1,2* 140 = 168

( d min base on table 4.3 page 128- C s ơ ở thi t k máy_Nguy n H u L c ) ế ế ễ ữ ộ

- Base on the standard :

• Choose driving pully diameter d1 = 180mm

1.2 Preliminary determine Center distance (a)

• The center distance a is determine through :

2( d1+d2 ) a 0,55( d≥ ≥ 1+d2 ) + h

 2(180+450) ≥ ≥ a 0.55(180+450) + 10.5

 1260 ≥ ≥ a 357

• The preliminary center distance a : asb = d2 = 450 = 450

1.3 Specifically Calculate Belt Length

• Belt length :  = 2 + y(V 0 3V )

' +(V0 lV ) 0

IT

= (2 ∗ 450) + y( I5;3#B; )

' +( I5;l#B; ) 0 ( I∗I5; ) = 1930,1 vv

( At standard we choose L= 2000) (table4.3 page 128 – c s thi t k ơ ở ế ếmáy_Nguyễn H u L c ) ữ ộ

• Check the revolution per sec of belt follow the equation :

^ = O = 5.482 = 2.74 !l#≤ }^~ = 10!l#

• The center distance a :

 = C + √C4'− 8∆'With:

Cv : Coefficient about speed effect

…@= 1 − 0.05 0.01O( '− 1 = 0.9849)

Cu : Coefficient of gear ratio u

C u = 1,14 ( base on table 4.9 page 152- C s ơ ởthiết k máy_ Nguy n H u L c) ế ễ ữ ộ

CL : Coefficent taking account of belt length

…‡= +s; = +s 20002240= 0.981

Follow Pics 4.21 page 151- C s thi t k máy_ Nguy n H u L c We ơ ở ế ế ễ ữ ộchoose L0 of Belt type B is 2240mm

CZ : Coefficent about number of belts , we choose = 0.95 (z=(2:3))

Cr : =0,9

( table 4.8 page 148- C s thi t k máy_ Nguy n H u L c ) ơ ở ế ế ễ ữ ộ

[P0] = 6 when d = 180 , L0 = 2240 and v = 5,48 m/s and is type B Belt From all the information we can easily calculate Z :

ˆ ≥ 1.44 we choose Z = 2 1.5: Specification of pulley

Based on the number of belts, we can determine the pulley width, following equation

Initial force applies on 1 belt, based on equation 4.19 [1]

Š;=780>O…#‹đ

†„ + Š@=13.71 × 0.89 × 2 + 33.47780 × 7.5 × 1.25

= 333.11(u) Which all the factor is calculated and described as

‹đ = 1.25 based on table 4.7 [1]

Š@= •SO'= 0.178 × 13.71'= 33.47 u based on equation 4.22 [1]

Useful force of the belt

Š2=1000>„O =# 1000 × 7.52 × 13.71= 273.5(u)

Tension force in the tight and lack side is demonstrated, followed equation 4.14 [1]

Š#= Š;+Š2

2 = 171.16 +122.802 = 227.55 (u) Š'= Š;−Š22 = 171.16 −122.802 = 114.75 (u)

1.7 Recalculating about the lifespan of the belt transmission

The maximum stress of the belt is figured out by the equation

ŽSTU= Ž#+ Ž@+ Že# = Ž;+ 0.5Ž2+ Ž@+ Že#

With :

ŽSTU=Š;

• +2• +Š2 Š@• +2P;_#• =333.05138 + 273.462 × 138 +33.47138+2 × 4276 × 100

ŽSTU= 8.09 (MPa) The life span of the belt

2 Design and Calculate Helical Gears System:

2.1 Calculating and reinforce calculation of Spur Gear

Determine all the kind of stress:

Base on the table 6.2 [1] we have,

Limitation of contact stress ŽšL›S;

Second torque loaded R' 0.7R u vv

Third torque loaded R) 0.9R u vv

Time interval of R# # 2315 ℎ !

Time interval of R' ' 4014 ℎ !

Time interval of R) ) 2470 ℎ !

ŽšL›S'; = 2—‰'+ 70 = 2 × 230 + 70 = 530 –>( ) Limitation of bending stress Ž?L›S;

Ž?L›S#; = 1.8 × 245 = 441 –>( ) Ž?L›S'; = 1.8 × 230 = 414 –>( ) Allowable overload stress

Base on equation (6.10) [1] and equation (6.11) [1]

}Žš~STU= 2.8 × 450 = 1260 –>( ) }Ž?~STU#= 0.8 × 580 = 464 –>( ) }Ž?~STU'= 0.8 × 450 = 360 –>( ) Allowable contact stress, based on equation (6.1), (6.3), (6.5) and (6.7) [1]

}Žš~#= ŽšL›S#o ׋š‡#

•š# = 560 ×11.1 = 509.09 (–>) }Žš~'= ŽšL›S'o ׋š‡'2 = 530 ×11.1 = 481.82(–>)

So, when we in the calculating process we take the value of the contact stress equal:

}Žš~ = v^K Žž} š~#, }Ž š~'Ÿ = 481.82 (–>) All the factor which is used in the 2 equations above, can be calculated in the process below

•š#= •š'= 1.1 gained by the table 6.2 [1]

uš # = 30—‰#'.I= 30 × 245'.I= 16259974.4

uš¡#= 60E ∑ ,Y£

Y w¤¥/)K››

uš¡#= 60 × 1 × 582 × (1 × 2315 + 0.7( ))× 4014 + 0.9( ))× 2470) uš¡#= 191795725.4

Because of uš¡#> uš #, we conclude that lifespan factor ‹š‡#= 1 Similarity with the process above, we can obtain that:

uš ' = 13972305.13

uš¡'= 27190604.35

So that, based on the condition uš¡'> uš ', ‹š‡'= 1

Allowable bending stress, based on equation (6.2), (6.4) and (6.8) [1]

}Ž?~#= Ž?L›S#o ׋?‡#

•?# = 441 ×

11.75 = 252(–>) }Ž?~'= Ž?L›S'o ׋?‡'•?' = 414 ×1.75 = 236.61 (–>)

All the factor which is used in the 2 equations above, can be calculated in the process below

Because of the u?¡#> u? , we conclude that lifespan factor ‹?¦#= 1

Similarity with the process above, we can obtain that:

u? = 4 × 108

u?¡'= 27190604.35

So that, based on the condition u?¡'> u? , ‹?¦'= 1

After calculating preliminary contact, bending and overload stress, we conclude all the data in the table below

Material of the Spur gear system

Element Material treatment Heat Hardness Ž˜ Ž™“ }Žš~ }Ž?~

–>

2.1.2 Design and reinforce calculation of the Spur gear

§= ‹T( ± 1)+ }ŽR#‹š©

š~' Ψ˜T 4

§= 43 4 + 1( )+ 115744.52 × 1.124(495.45)'× 4 × 0.25= 173.78 vv( ) All the factor is selected and figure out in the process bellow

‹T= 49.5 base on table (6.5) [1]

Ψ˜T= 0.3 base on standard and table 6.6 [1]

Ψ˜V= 0.53Ψ˜T( + 1) = 0.53 × 0.3 × (4 + 1) = 0.6625 base on equation (6.16) [1]

‹š©= 1.12 is selected base on table 6.7, diagram 3 [1]

We choose the center distance base on standard SEV229-75, §= 200 mm( )

Determine the meshing parameter of the spur gears

Following equation (6.17) [1]:

vW= (0.01 ÷ 0.02 ) §= (0.01 ÷ 0.02 200 = 2 ÷ 4) ( )

We choose the normal module of spur gear vW= 3

The width of the teeth

§'= Ψ˜T§= 0.3 × 200 = 60 vv

§#= §'+ 5 = 60 + 5 = 65mm The number of teeth of the driving spur gear

2 ∗  § ∗ cos (8)

v W ∗ ( + 1) ≥ „ # ≥2 ∗ § ∗ cos (20)

v W ∗ ( + 1) 26.4 ≥ „ # ≥ 25.05

We choose „#= 26

The number of teeth of the driven spur gear

„'= „#= 4 × 26 = 104

We choose „'= 104

We figure out the error in transmitting ratio

p¯TL=„#„'=10426= 4The error of the transmitting ratio base on the calculating ratio

° = ± − p¯TL

p¯TL ± × 100% = 0% ≪ 5% ⟹ •EEœ_

So, we recalculate the center distance and the ration transmission

§=v(„#2+ „')=3(26 + 104)2 = 195 (vv) Calculate the lead angle

µ = EE!v ∗ „22§ = 12.83 3.2.3 Reinforce calculation of strength of the spur gear

Reinforce calculation of contact stress on the system

Base on the table (6.5) [1] we choose the mechanical factor of the material

ˆ¶= 274 (–>)#/)Without adjusting design, we can conclude:

Angle profile of the tooth equals the meshing angle

‚2= ‚2§= tanl#·cos µ¸ = tantan ‚ l#¹tan 200.975 º = 20.47;

Base on the equation (6.35) [1], we obtain that angle of the tooth on the unit cylinder

µ˜= tanl#(cos‚2tan µ = tan) l#( (cos 20.47) tan 12.839 ≈ 12.052( ))

The factor of the shape of the contact area ˆš base on the equation (6.34) [1]

ˆš= +sin 2 ‚2§2 cos µ˜ = +sin 2 × 20.4712 cos(12.052)( ) = 1.7277

°†= »1.88 − 3.2 ·„#1 +„1

'¸ cos µ = 1.88 − 3.2 ·¼ » 26 +1 104¸¼ 0.975 = 1.731

The factor of vertical meshing of the gear °©, based on equation (6.37) [1]

°©= §sin µ½vW= ¾˜T§sin µ½vW= 0.25 × 200 ×sin(12.83)3½ = 1.178 Because of the factor °©= 1.178, the factor of tooth meshing ˆ¿, based on the equation (6.36a) [1]

ˆ¿= +°†1 = +1.73 = 0.761Pitch circle of the pinion _§#

_§#= + 1 =2§ 2 × 2004 + 1 = 80(vv) Linear velocity of the pinion

O#=½_§#K#

60000 =½ × 80 × 58260000 = 2.44 (v/!) Because of O#= 2.44 (v/!), based on the table (6.13) [1], we choose degree of

accuracy is 9 Therefore, from the table (6.14), the factor of load uniformly

distributes ‹š†= 1.13

The factor Oš is calculated, following equation (6.42)

Oš= ŽšoO#:§ = 0.002 × 73 × 2.44 ×+200

4 = 2.52With Žš and o is the factor which can be gain in the table 6.15 and 6.16 in [1], respectively

‹š@= 1 +Oš2R§_§##‹š©‹š†∗ §= 1 +2 × 115744.52 × 1.12 × 1.13 ≈ 1.032.52 × 0.25 × 200 ∗ 80

Base on equation (6.39) [1], we gain the total load factor when we consider

contact stress ‹š.

‹š= ‹š©‹š†‹š@= 1.12 × 1.13 × 1.04 = 1.31

From all the factor we calculating, we can estimate the contact stress in the

Žš= ˆ¶ˆšˆ¿:2R#‹š( + 1)(§ _§#' )

Žš= 274 × 1.74 × 0.77 × 2 × 115744 × 1.32 × 5.23 + 1 × 50 × 64.12À ( ) '

Žš= 380.6 After we estimate the contact stress appear on the system, we continue

calculating exactly the allowable contact stress }Žš~

Base on equation (6.1), we get

}Žš~ = }Žš~ × ˆ@× ˆÁ× ‹Uš= 495.45 × 1 × 0.95 × 1 = 470,7 (–>) With selecting all the factor is demonstrated in the process below

ˆ@= 1, because O#= 2.42 < 5 (v/!)

Manufacturing with ÃT= 1.6 Äv ⟺ ˆÁ= 0.95, based on table 5.5 [4]

‹Uš= 1

}Žš~ = 495.45

Therefore, Žš= 375.62 ≤ Ž} š~ = 470.7 So, we can conclude that we satisfy

the contact stress condition.

Reinforce calculation of bending stress on the system

From the table 6.7 and table 6.14 [1], we get: ÆO#< 2.5 v/! ⟺ ‹?†‹?©= 1.24 = 1.37The factor O? can demonstrated, following equation (6.47) [1]

O?= Ž?oO#:§ = 0.006 × 73 × 2.44 ×+2004 = 7.55

With the factor Ž? and ; can be obtain in the table 6.15 and 6.16 in [1]

So that, the factor of the kinematic load appears in meshing zone when we calculate the bending stress:

‹?@= 1 +O2R?§_§##‹?©‹?†= 1 +2 × 115744.52 × 1.24 × 1.37 = 1.087.55 × 0.25 × 200 × 80

The factor of load when we consider bending stress

‹?= ‹?©‹?†‹?@= 1.24 × 1.37 × 1.09 = 1.85 Base on equation (6.43) [1]

⎩

⎨

⎧ Ê¿=°†1 =1.70 = 0.591

Ê© = 1 −µ140 = 1 −10.735140 = 0.92The factor of the shape of the tooth Ê?, which is depend on the equivalent tooth

„@ and the table 6.18 [1]

Æ Ê?#= 4Ê?'= 3.6

equation (6.43) [1]

For the pinion

Ž?#=2R#‹?Ê§_§#v =¿Ê©Ê?# 2 × 115744 × 1.85 × 0.59 × 0.92 × 40.25 × 200 × 80 × 3 = 74.11(–>) For the driven gear

Ž?'= Ž?#×Ê?'

Ê?#= 74.11 ×3.64 = 69.7(–>)

Determine the exactly allowable bending stress in the system

Base on the equation (6.2) [1]

}Ž?#~ = }Ž?#~ÊÁÊË‹U?= 252 × 1 × 1 × 1 = 252 –>( )

}Ž?'~ = }Ž?'~ÊÁÊË‹U?= 237.43 × 1 × 1 × 1 = 237.43 –>( )

We can obtain that:

Æ Ž?#= 57.75 –>( ) < }Ž?#~ = 252(–>)Ž?'= 51.97 –>( ) < }Ž?'~ = 237.43(–>)Therefore, we satisfy the bending stress condition

Reinforce the overload stress.

Base on [2] we obtain that the factor of overload ‹Ì2= 3.2

Determine the maximum contact stress ŽšSTU

ŽšSTU= ŽšÀ‹Ì2= 470,7 3.2 = 841.98 –>√ ( ) < }Žš~STU= 1260(–>) Determine the maximum bending stress Ž?STU

ÍŽ?STU#= Ž?#À‹Ì2= 74.11 3.2 = 237.15 < Ž√ } ?STU#~ = 464(–>)Ž?STU'= Ž?'À‹Ì2= 66.7 3.2 = 213.43 < Ž√ } ?STU'~ = 360(–>)Therefore, we satisfy the overload stress condition

In conclusion, we successfully reinforce the strength of the Helical Gear System 2.2.4 Force calculating of the helical gear transmission system

Useful Force is determined by the equation (6.16) [2]

Š2#=_§#2R#=2 × 115744.5280 = 2893.61(u) = Š 2'

Radial Force is calculated from the equation in (6.17) [2]

Šp#= Šp'=Š2#tan ‚W§cos µ =2893.61 tan 20cos 12.839 = 1107.89 (u)

Axial force is described in the equation (6.18) [2]

ŠT#= Š2#tan µ = 2893.61 tan 12.839 = 659.46(u)

Chapter 4: Design and Calculate Spur Gear System

Input data of the spur gear

Power transmitting > 6.84 CD

Second torque loaded R' 0.7R u vv

Third torque loaded R) 0.9R u vv

Determine all the kind of stress:

Base on the table 6.2 [1] we have,

Limitation of contact stress ŽšL›S;

ŽšL›S#; = 2—‰#+ 70 = 2 × 245 + 70 = 560 –>( ) ŽšL›S'; = 2—‰'+ 70 = 2 × 230 + 70 = 530 –>( ) Limitation of bending stress Ž?L›S;

Ž?L›S#; = 1.8 × 245 = 441 –>( )

Ž?L›S'; = 1.8 × 230 = 414 –>( ) Allowable overload stress

Base on equation (6.10) [1] and equation (6.11) [1]

}Žš~STU= 2.8 × 450 = 1260 –>( ) }Ž?~STU#= 0.8 × 580 = 464 –>( ) }Ž?~STU'= 0.8 × 450 = 360 –>( ) Allowable contact stress, based on equation (6.1), (6.3), (6.5) and (6.7) [1]

}Žš~#= ŽšL›S#o ׋š‡#•š# = 560 ×11.1 = 509.09 (–>) }Žš~'= ŽšL›S'o ׋š‡'2 = 530 ×11.1 = 481.82(–>)

So, when we in the calculating process we take the value of the contact stress equal:

}Žš~ = v^K Žž} š~#, }Žš~'Ÿ = 481.82 (–>) All the factor which is used in the 2 equations above, can be calculated in the process below

•š#= •š'= 1.1 gained by the table 6.2 [1]

uš # = 30—‰#'.I= 30 × 245'.I= 16259974.4

uš¡#= 60E ∑ ,Y£

Y w¤¥/)K››

uš¡#= 60 × 1 × 145.5 × (1 × 2315 + 0.7( ))× 4014 + 0.9( ))× 2470) uš¡#= 47948931

Because of uš¡#> uš #, we conclude that lifespan factor ‹š‡#= 1 Similarity with the process above, we can obtain that:

uš ' = 13972305.13

uš¡'= 15982977.12

So that, based on the condition uš¡'> uš ', ‹š‡'= 1

Allowable bending stress, based on equation (6.2), (6.4) and (6.8) [1]

}Ž?~#= Ž?L›S#o ׋?‡#

•?# = 441 ×

11.75 = 252(–>) }Ž?~'= Ž?L›S'o ׋?‡'•?' = 414 ×1.75 = 236.61 (–>)

All the factor which is used in the 2 equations above, can be calculated in the process below

Because of the u?¡#> u? , we conclude that lifespan factor ‹?¦#= 1

Similarity with the process above, we can obtain that:

u? = 4 × 108

u?¡'= 11930715.86

So that, based on the condition u?¡'> u? , ‹?¦'= 1

After calculating preliminary contact, bending and overload stress, we conclude all the data in the table below

Table : Material of the Spur gear system

Element Material treatment Heat Hardness Ž˜ Ž™“ }Žš~ }Ž?~

–>

4.1.2 Design and reinforce calculation of the Spur gear

§= ‹T( ± 1)+ }ŽR#‹š©

š~' Ψ˜T 4

§= 49.5 3 + 1( )+ 449181.35 × 1.074(495.45)'× 3 × 0.3= 261.4 vv( ) All the factor is selected and figure out in the process bellow

‹T= 49.5 base on table (6.5) [1]

Ψ˜T= 0.3 base on standard and table 6.6 [1]

Ψ˜V= 0.53Ψ˜T( + 1) = 0.53 × 0.3 × (3.06 + 1) = 0.646 base on equation (6.16) [1]

‹š©= 1.07 is selected base on table 6.7, diagram 3 [1]

We choose the center distance base on standard SEV229-75, §= 250 mm( )

Determine the meshing parameter of the spur gears

Following equation (6.17) [1]:

vW= (0.01 ÷ 0.02 ) §= (0.01 ÷ 0.02 250 = 2.5 ÷ 5) ( )

We choose the normal module of spur gear vW= 3

The width of the teeth

§'= Ψ˜T§= 0.3 × 250 = 75 vv

§#= §'+ 5 = 75 + 5 = 80 mm The number of teeth of the driving spur gear

So, we recalculate the center distance and the ration transmission

§=v(„#+ „')

2 =3(41 + 126)2 = 250.5 (vv) Because of the number of teeth of the driving gear „#> 30, so the center distance is assured

S=„„'

#=12641= 3.07

⇒ ∆ = − S× 100% = ±3.06 − 3.073.06 ± × 100% = 0.3%

So, the deviation is smaller than 5% is this ratio is accpected

4.1.3 Reinforce calculation of strength of the spur gear

Reinforce calculation of contact stress on the system

Base on the table (6.5) [1] we choose the mechanical factor of the material

ˆ¶= 274 (–>)#/)Without adjusting design, we can conclude:

Angle profile of the tooth equals the meshing angle

‚2= ‚2§= 20;

The factor of the shape of the contact area ˆš base on the equation (6.34) [1]

ˆš= +sin 2 ‚2§2 cos µ˜ = +sin 2 × 202 cos(0)( ) = 1.764

ˆ¿= +(4 − °3 †)= +(4 − 1.70)3 = 0.861 Pitch circle of the pinion _§#

_§#= 2§ + 1 =2 × 2503.07 + 1 = 122.75(vv) Linear velocity of the pinion

O#=½_60000 =§#K# ½ × 122.75 × 145.560000 = 0.94 (v/!)

Because of O#= 0.89 (v/!), based on the table (6.13) [1], we choose degree of

accuracy is 9 Therefore, from the table (6.14), the factor of load uniformly

From all the factor we calculating, we can estimate the contact stress in the

Žš= ˆ¶ˆšˆ¿:2R#‹š( + 1)(§_§#' )

Žš= 274 × 1.764 × 0.861 × 2 × 449181.35 × 1.32 × 3 + 1 × 80 × 64.12À ( ) '

Žš= 476.92 (–>)

After we estimate the contact stress appear on the system, we continue

calculating exactly the allowable contact stress }Žš~

Base on equation (6.1), we get

}Žš~ = }Žš~ × ˆ@× ˆÁ× ‹Uš= 481.82 × 1 × 0.95 × 1 = 457.73 (–>) With selecting all the factor is demonstrated in the process below

ˆ@= 1, because O#= 0.89 < 5 (v/!)

Manufacturing with ÃT= 1.6 Äv ⟺ ˆÁ= 0.95, based on table 5.5 [4]

‹Uš= 1

}Žš~ = 481.82

Therefore, Žš= 457.73 ≤ Ž} š~ = 457.73 So, we can conclude that we satisfy

the contact stress condition.

Reinforce calculation of bending stress on the system

From the table 6.7 and table 6.14 [1], we get: ÆO#< 0.89 v/! ⟺ ‹?†‹?©= 1.17 = 1.13The factor O? can demonstrated, following equation (6.47) [1]

O?= Ž?oO#:§

= 0.016 × 73 × 0.94 ×+2503 = 9.85With the factor Ž? and ; can be obtain in the table 6.15 and 6.16 in [1]

So that, the factor of the kinematic load appears in meshing zone when we calculate the bending stress:

‹?@= 1 +O ?§_§#

2R#‹?©‹?†= 1 +2 × 449181.35 × 1.24 × 1.37 = 1.089.85 × 0.25 × 250 × 122.75

‹?= ‹?©‹?†‹?@= 1.17 × 1.13 × 1.08 = 1.42 Base on equation (6.43) [1]

The factor of the shape of the tooth Ê?, which is depend on the equivalent tooth

„@ and the table 6.18 [1]

ÆÊ?#= 3.7Ê?'= 3.6

equation (6.43) [1]

For the pinion

Ž?#=2R#‹?Ê¿Ê©Ê?#

§_§#v =2 × 449181.35 × 1.42 × 0.56 × 1 × 3.70.3 × 250 × 122.75 × 3 = 96.4(–>) For the driven gear

Ž?'= Ž?#×ÊÊ?'

?#= 96.4 ×3.63.7 = 93.8(–>)

Determine the exactly allowable bending stress in the system

Base on the equation (6.2) [1]

}Ž?#~ = }Ž?#~ÊÁÊ Ë‹U?= 252 × 1 × 1 × 0.95 = 240.27 –>( ) }Ž?'~ = }Ž?'~ÊÁÊ Ë‹U?= 237.43 × 1 × 1 × 0.95 = 225.56 –>( )

We can obtain that:

ÆŽ?#= 76.43 –>( ) < }Ž?#~ = 240.27(–>)Ž?'= 74.37 –>( ) < }Ž?'~ = 225.56(–>)Therefore, we satisfy the bending stress condition

Reinforce the overload stress.

Base on [2] we obtain that the factor of overload ‹Ì2= 3.2

Determine the maximum contact stress ŽšSTU

ŽšSTU= ŽšÀ‹Ì2= 476.92 3.2 = 853.13 –>√ ( ) < }Žš~STU= 1260(–>) Determine the maximum bending stress Ž?STU

ÑŽ?STU#= Ž?#‹Ì2= 96.40 × 3.2 = 308.48 < Ž} ?STU#~ = 464(–>)

Ž?STU'= Ž?'‹Ì2= 93.8 × 3.2 = 300.14 < Ž} ?STU'~ = 360(–>)Therefore, we satisfy the overload stress condition

In conclusion, we successfully reinforce the strength of the Spur Gear System

4.2.4 Force calculating of the Spur gear

Useful force, equation 6.13[2]

Š2#= Š2'=2R#

_§#=2 × 449181.35122.75 = 7318.37(u) Radial Force, equation 6.14 [2]

Šp#= Šp'= Š2#tan ‚§= 7318.37 tan 20 = 2663.67 (u)