đồ án hệ thống truyền động design tramission system for truck on crane rail

Fatigue limit stressesThe limitation of contact stress of driving gear from the Table 6.2 [1]σHlim 12.. Safety coefficientsSSafety coefficientsSfor contact stress based on Table 6.2 [1]:

HO CHI MINH UNIVERSITY OF TECHNOLOGY FALCUTY OF MECHANICAL ENGINEERING

TRANSMISSION SYSTEM PROJECTGroup CC01 – Project 1 – Data 1

DESIGN TRAMISSION SYSTEM FOR TRUCK ON CRANE RAIL

SEMESTER 232 SCHOOL YEAR 2023-2024

Instructor: Assoc Prof Nguyễn Tấn Tiến

Nowadays, we live in Industry 4.0 which based on a digital platform and integrating smart device to optimize production technology Applying technology to manufacture not only increase productivities, but also a safety workplace for humans So all we need now it is very important to us to develop and improve our technical skills from the beginning of our university.Mechanical engineering is study of physics, mathematics and material science to design, analyze the kinematics and dynamics of a system, manufacture and maintain a mechanical system It is foundation that many industries’ and product as car, plane, etc All of that need a transmission system

In this transmission project, we design a mechanical system that transmit motion from the motor to the truck on the crane rail During the period of time doing this project, we can reinforce all the knowledge from courses Kinematics and Dynamics of Machine, Machine Elements, Mechanical Engineering Drawing Also, we improve the skills need for using engineering software such as CAD and Solid Work With the instruction of Pro Nguyễn Tấn Tiến

Rail resistance force, F=3860(N )

Tangential velocity on the wheel, v=1.6 (m/s)

Diameter of the wheel, D=330 (mm)

Service life, Lh=4 years=19200hours

One-direction working, 2 shifts per working day, 300 working days/ year, hour shift

8-CHAPTER 1: SELECTING THE MOTOR AND SPEED RATIO DISTRIBUTION 7

I Transmission system diagram 7

II Crane rail wheel parameters 7

II Choosing the efficiency of the transmitting system 7

III Choosing the motor 7

V Choosing transmission ratio 8

VI Speed of shaft 8

VII Power of shafts: 8

VIII Torque of shafts 9

IX Summary 9

CHAPTER 2: DESIGN THE SPUR GEAR FOR THE SPEED REDUCER 10

I Choosing the material 10

II Allowable stresses 10

1 Fatigue limit stresses 10

2 Theorical operating cycles 10

3 Equivalent operating cycles 10

4 Lifespan factor 10

5 Safety coefficients 10

6 Allowable contact stress 10

7 Allowable bending stress 11

8 Allowable overload stress 11

9 Summary 11

III Center distance 11

IV Gear module 11

V Teeth of gears 11

VI Face width 12

VII Accuracy class 12

VII Verify contact stess 12

IX Verify bending stress 13

X Force caculating of the Spur Gear 14

XI Parameters of gear drive 14

CHAPTER 3: DESIGN THE SPUR GEAR FOR THE TRANSMISSION 15

I Choosing the material 15

II Allowable stresses 15

1 Fatigue limit stresses 15

2 Theorical operating cycles 15

3 Equivalent operating cycles 15

4 Lifespan factor 15

5 Safety coefficients 15

6 Allowable contact stress 15

7 Allowable bending stress 16

8 Allowable overload stress 16

9 Summary 16

III Center distance 16

IV Gear module 16

V Teeth of gears 16

VI Adjusting the gear profile 17

VII Gear parameters 17

VIII Face width 18

VII Accuracy class 18

IX Verify contact stess 18

X Verify bending stress 19

XI Force caculating of the Spur Gear 20

XII Summary 20

II Determine forces acting on the shaft 21

II Determine and caculating Shaft I 21

1 Determine the preliminary diameter of shaft I 21

2 Caculating the distance between the sections of shaft I 21

3 Caculate force and moment apply on the shaft I 22

4 Determine diamter of each section and key: 24

III Determine and caculating Shaft II 24

1 Determine the preliminary diameter of shaft II 24

2 Caculating the distance between the sections of shaft II 25

3 Caculate force and moment apply on the shaft II 25

4 Determine diamter of each section and key 26

IV Re-caculating Shaft and Key 27

1 Re-caculaing shaft condisdering fatigue strength 27

2 Re-caculate strength condition of keyway 29

CHAPTER 6: DESIGN AND CACULATE ROLLING BEARING 30

I Select bearings for shaft I 30

1 Choose type of bearing 30

2 Caculate the dynamic load 30

II Select bearings for shaft II 30

1 Choose type of bearing 30

2 Caculate the dynamic load 31

CHAPTER 5: DESIGN FLEXIBLE COUPLING 32

I Specification of flexible coupling 32

II Caculate force apply on the flexible coupling 32

III Strength of the coupling 32

CHAPTER 8: LUBRICATION 33

II.Lubrication condition 33

II Select the type and of oil 33

CHAPTER 9: TOLERANCE AND ASSEMBLY TYPE SELECTION 34

I Tolerance of key and key-way 34

II Tolerance of spur gears 34

III Tolerance assembly of bearing 34

IV Tolerance assembly of oil seal 34

V Tolerance assembly of caps of hub 34

I Transmission system diagram

Figure 1 Truck on crane rail system

II Crane rail wheel parameters

The power of the wheel using the formula (2.11) [1]:

Plv= Fv

1000=

3860 1.6×

1000 =6.176(kW)The speed of the wheel using the formula (2.16) [1]:

nwheel=60000 v

Dπ =

60000 ×1.6

330 ×π =92.6(rpm)

II Choosing the efficiency of the transmitting system

The efficiency of each machine element is chosen based on the Table 2.3 [1]

Table 1.1 Efficiency of system

Transmission parts Symbols Efficiency factor

Therefore, the general efficiency of the transmitting system using formula (2.9) [1]:

ηsys=ηsnηsrηrg ηsgηcg=0.93 × 0.96 ×0.99×0.98 0.98× ≈0.84

III Choosing the motor

The general ratio of transmission is defined in the Table 2.4 [1]:

ugen=ugearreducerugeartrans= (3 ÷ 5) × ( 4 ÷ 6)= (12 30÷ )

The primality rotation speed of motor is defined by the formula (2.18) [1]:

Current(A)

Powerfactor

Efficiency(%)

Torque(Nm)

NetWeigh

t (kg)

3 GBP132323− 7.5 1450 14.5 0.81 89.3 49.4 73

V Choosing transmission ratio.

From formula (2.18) in [1], recalculate the general transmission ratio of the system:

∆u=|3.915 4− |

4 =2.125% <[∆u]=4 %

→ Allowable gear transmission ratio error

Note: [∆u]=4 % Allowable ratio error based on page 99 [1]

VI Speed of shaft.

Speed of shaft I:

n n= =1450(rpm)

Speed of shaft II:

VII Power of shafts:

Power of working part:

PI= PII

ηbg×ηsr

=0.99 ×0.966.845 =7.202(kW)Power of motor shaft:

TII=9.55 ×106

× II

nII=9.55 ×106

×6.845362.5=180330.35(Nmm)Torque of wheel:

Transmission

Torque, Nm 48.41 47.43 180.03 636.94

I Choosing the material

We choose C 45 steel with structure improvement method for designing the gear drive (following the Table 6.1) According to the table, we consider group I which has the hardness HB1=260 HB for the driving gear For the driven gear, we should choose hardness according to the formula HB1≥H B2+ ( 10 ÷15)HB So we choose HB2=245 HB

II Allowable stresses

1 Fatigue limit stresses

The limitation of contact stress of driving gear from the Table 6.2 [1]

2 Theoretical operating cycles

Driving gear’s contact operating cycles from formula (6.5) [1]:

NFO1=NFO2=4 × 106

(cycles)

3 Equivalent operating cycles

Driving gear’s equivalent operating cycles from formula (6.7) [1]:

NHE1=NFE1=60 nILhc=60 ×1450 ×19200 ×1=1,67 10× 9(cycles)

Driven gear’s equivalent operating cycles from formula (6.7) [1]:

NHE2=NFE2=60 nIILhc=60 × 362.5× 19200×1=4,176 10× 8

(cycles)

4 Lifespan factor

Driving gear’s life factor for contact stress:KHL1=1because NHE1>NH 01

Driven gear’s life factor for contact stress:KHL 2=1because NHE1>NH 01

Driving gear’s life factor for bending stress:KFL 1=1because NHE1>NH 01Driven gear’s life factor for bending stress:KFL 2=1 because NHE1>NH 01

5 Safety coefficientsS

Safety coefficientsSfor contact stress based on Table 6.2 [1]: SH=1.1Safety coefficientsSfor bending stress based on Table 6.2 [1]: SF=1.75

6 Allowable contact stress

Allowable contact stress for driving gear based on formula (6.1a) [1]

[σH 1]=σHlim 10

KHL 1

SH

=590 ×1.11 =536.36(MPa ) Allowable contact stress for driven gear based on formula (6.1a) [1][σH 2]=σHlim 20 KHL 2

SH=560 × 1

1.1=509.09 (MPa ) Allowable contact stress for calculation based on page 95 [1]

[σH]=min{ [σH1],[σH 2] }=509.09(MPa )

7 Allowable bending stress

Allowable bending stress for driving gear based on formula (6.1a) [1][σF 1]=σFlim 1

0 KFL1

SF

=468 × 11.75=267.43( MPa )Allowable contact stress for driven gear based on formula (6.1a) [1][σF 2]=σFlim 20 KFL2

SF

=441× 11.75=252( MPa) Allowable contact stress for calculation based on page 95 [1]

[σF]=min{ [σF 1],[σF 2] }=252( MPa)

8 Allowable overload stress

Allowable overload contact stress based on formula (6.13) [1]

[σH]max=2.8 σb=2.8 ×650 1820= (MPa)

Allowable overload bending stress based on formula (6.14) [1]

[σF 1]max=0.8 σb=0.8 ×650 520 ( MPa ) =

9 Summary

Table 2.1 Properties of Spur gear material

nt

Hardness

Strengt

h Stress

σb,MPa

YieldStress

σc h,MPa

[σH]MPa

[σF]MPaDrivin

improves

Ka=49.5 based on the Table 6.4 [1]

ψba=0.4

ψbd=0.53ψba( u+1)=0.53× 0.4 ×5=1.06 based on formula (6.16) [1]

KHβ=1.05 based on the Table 6.7 [1]

1 Determine the preliminary diameter of shaft I

Based on the equation (10.9) [1], we figure out the preliminary diameter

[τ] – Allowable torsional stress, [τ]=15 ÷ 30 MPa according to page 188 in [1]

[τ]=20 MPa – for shaft I and shaft II

We choose the preliminary diameter of shaft I based on the Table 10.2 [1]

dI=25 mm with the width of bearing b01=16 mm

2 Calculating the distance between the sections of shaft I

Following Table 10.3 and 10.4 [1] and equation 10.10, we figure out all the distance of shaft I:

The mayo length of half of coupling:

lm12=(1.4 2.5÷ ) dI=( 35 62.5÷ )(mm)

We choose lm12=50 mm based on Table 16.10a and 16.10b [2]

The mayo length of the driving gear:

3 Calculate force and moment apply on the shaft I

We consider the magnitude of forces acting on shaft I on the table below:

Table 4.2 Force acting on the shaft I

Magnitude, N 1897.36 690.58 340.9

According to the free body diagram in figure, we have force equation equilibrium and moment equation of equilibrium in the Oyz surface as:

Table 4.3 Reaction force forces of bearing

Magnitude, N 1161.32 345.29 395.14 345.29After getting the value of forces, we draw the moment diagram for the shaft I

Based on the moment diagram, we calculate the bending moment Mjand quivalent moment Mjand equivalent moment Meqj at the section j along the length of shaft I by equation (10.15) and (10.16)

Meqj=√Myj

2

+Mxj 2

+0.75Tj 2

Table 4.4 Specification of key for shaft I

III Determine and calculating Shaft II

1 Determine the preliminary diameter of shaft II

Based on the equation (10.9) [1], we figure out the preliminary diameter

of the driven gear of reducer

dII=3√ TII

0.2[τ]=

3

√180330.350.2× 20 =35.6(mm)Where,

[τ] – Allowable torsional stress, [τ]=15 ÷ 30 MPa according to page 188 in [1]

[τ]=20 MPa – for shaft I and shaft II

We choose the preliminary diameter of shaft II based on the Table 10.2 [1]:

dII=40 mm with the width of bearing b02=23 mm

2 Calculating the distance between the sections of shaft II

Following Table 10.3 and 10.4 [1] and equation 10.10, we figure out all the distance of shaft II:

The mayo length of driving gear transmission:

3 Calculate force and moment apply on the shaft II

We consider the magnitude of forces acting on shaft I on the table below:

Table 4.5 Force acting on the shaft II

Force Gear of reducer Gear of transmission

Ft 2 Fr 2 Ft 3 Fr 3

Magnitude, N 1897.36 690.58 4434 1648.1According to the free body diagram in figure, we have force equation equilibrium and moment equation of equilibrium in the Oyz surface as:

{∑Fy 0(↓

¿ =− Fr 3+ FBy+ Fr 2− FDy=0¿∑MB( ↺¿¿− Fr 3×AB− Fr 2×BC+ FDy×BD=0¿

Table 4.6 Reaction force forces of bearing

Magnitude, N 8763.1 2559.3 2431.7 1601.76After getting the value of forces, we draw the moment diagram for the shaft II

Based on the moment diagram, we calculate the bending moment Mj

and quivalent moment Mjand equivalent moment Meqj at the section j along the length of shaft I by equation (10.15) and (10.16)

IV Recalculating Shaft and Key

1 Recalculating shaft considering fatigue strength

As we use steel C45 with σb=850 MPa We can get limitation of bending stress and torsion stress

σ−1=0.436 σb=0.436 × 850=370.6 MPa

τ−1=0.58 σ−1=0.58 ×370.6=214.95 MPa

According to Table 10.7 [1], we choose factor of bending stress ψσ=0.1and for torsional stress ψτ=0.05 We check the safety condition of shaft I and II by following task:

1 The bending resistance moment W and WO at the dangerous section Weuse formula from Table 10.6 [1] to calculate the value of W and WO

For 1 keyway shaft:

Table 4.10 Scale factor εσ and ετ

4 Calculate the value Kσd, Kτd and safety factor s

The value of Kσd and Kτd are determined by the formula (10.26) and (10.26):

So we can get the table below:

Table 4.11 Safety factors at dangerous section

Secti

on d ,mm

Kσ/εσ Kτ/ετ

Kσd Kτd sσ sτ sKeywa

y

Tensi

on onfix

Keyway

Tensi

on onfix

The condition of bending and shear strength of key according to formula(9.1) and (9.2) [1]

Length of key lt= (0.8 ÷ 0.9)lm , choose from Table 9.1a [1]

σd – Allowable bending stress strength, choose σd=150 MPa from Table 9.5 [1]

τc – Allowable shear stress strength, τc=( 60 90÷ ) MPa from Table 9.5 [1]Table 4.12 Key strength testing

I Specification of flexible coupling.

Based on the Table 16.10a and 16-10b [2] with the Torque T =48.4 Nm

We can choose flexible coupling have parameters below:

Table 5.1 Specification of flexible coupling

Paramet

3 dc

II Calculate force apply on the flexible coupling

Force applies on the coupling

Ft=2 Tmotor

Do

=2 × 48408.62

71 =1363.62(N)Force applies on the shaft

Fr=0.25 Ft=0.25× 1363.62 340.9= (N )

III Strength of the coupling

Condition for the crushing stress of flexible coupling

σd=2 kTmotor

ZDodcl3

=2× 1.4 × 48408.626 ×71 ×10 ×15 =2.12<[σd]=4 MPaWhere,

×6 =39.77 <[σd]=80 MPa

So the flexible coupling satisfies the stress condition

I Select bearings for shaft I

1 Choose type of bearing

Because the force applying to the spur gear So there is no Axial load,

Fa=0 N So we use the Radial Bearing

We calculate the radial force acting on bearing at A

=√395.33 345.29 524.752

= (N )Because the radial force at A is greater than the radial force at C So we use the radial force at C for calculating

We preliminary select bearing based on the diameter of the shaft and from SKF:

Table 6.1 Specification of 206 bearing from SKF

2 Calculate the dynamic load

Consider formula (11.3), dynamic load for the radial bearing with Fa=0 N :

Q= XVFrAktkd=1 × ×1 1211.6 ×1 ×1=1211.6N

Where,

V – Factor of rotating ring, choose V =1 for rotating inner

kđ – Service factor, choose kd=1

kt –Temperature factor, choose kt=1

Next, we calculate the loading capacityCd, using formula (11.1) [1]:

Cd=Qm√L=1211.6√1670.4 14.23 = (kN)

Where,

m – Degree of curvature, choose m=3 for the radial bearing

L – Longevity of bearing in million revolutions, based on formula (11.2)L=60 Lhn

106 =60 ×19200 ×1450

106 =1670.4(mil rev)The loading capacity Cd=14.4 kN<C=20.9 kN So this bearing satisfies the condition of dynamic load

II Select bearings for shaft II

1 Choose type of bearing

Because the force applying to the spur gear So the Axial load Fa=0 N

So we use the Radial Bearing

Calculate the radial force acting on bearing at A

F =√F 2+ F 2=√8646.6 25162+ 2=9129.14 (N )