ĐỒ án môn học CHI TIẾT máy THIẾT kế hệ dẫn ĐỘNG

Internal torque diagram of the second shaft II………..16 ABSTRACT The flexible power transmission elements comprises of belt drives, chain drives and rope drives.. For example let us consid

TRƯỜNG ĐẠI HỌC BÁCH KHOA HÀ NỘI

VIỆN CƠ KHÍ

BỘ MÔN CƠ SỞ THIẾT KẾ MÁY VÀ RÔ BỐT

ĐỒ ÁN MÔN HỌC

CHI TIẾT MÁY

ĐỀ:

Người hướng dẫn TS Đặng Bảo Lâm

Sinh viên thực hiện Đường Tiến Luận

Mã số sinh viên 20184997

Lớp chuyên ngành Cơ khí 04

Lớp tín chỉ

Ngày kí duyệt đồ án: ……./……./20… Ngày bảo vệ đồ án: ……./……./20…

Ký tên

CỦA THẦY HỎI THI

1

TABLE OF CONTENTS

Contents

ABSTRACT 4

INTRODUCTION 4

1 SELECTING THE ENGINE AND TRANSMISSION 4

2 DESIGNING THE CHAIN DRIVE 7

3 DESIGNING THE HELICAL GEARBOX 8

4 DESIGNING THE SHAFT 11

5 SELECTING AND ANALYZING THE BEARINGS 17

6 THE OUTSIDE STRUCTURES OF GEARBOX 18

7 CONCLUSION 18

8 REFERENCES 19

TABLE OF FIGURES

Fig 1: Distance between point of force application………12 Fig 2 Internal torque diagram of the first shaft (I)……… 13 Fig 3 Internal torque diagram of the second shaft (II)……… 16

ABSTRACT

The flexible power transmission elements comprises of belt drives, chain drives and rope drives Since these elements are flexible they are called flexible drives They are commonly used in short distance power transmission Wire ropes can also

be used for long distance power transmission Unlike rigid transmission elements(gears), the design of flexible drives are simple and reduces the cost When compared with rigid transmission elements, they are quiet and absorbs shocks which reduces mechanical vibrations

Flexible drives do not have infinite life, hence its necessary to consider the wear, aging, loss of elastic property and also the environment in which the drive is employed, while designing/selecting the flexible drives

INTRODUCTION

Transmission systems transfer mechanical power from a source to another machine components For example let us consider a car, in which the power from engine is transmitted to wheels through clutch, gearbox, prop shaft and differential, these components are called transmission elements If we consider a lathe, power from motor is provided to a chuck through belt drive and gearbox, these components are transmission elements A number of elements like gears in a gear box together form a transmission system

1 SELECTING THE ENGINE AND

TRANSMISSION

In mechanical drive systems, electric motors are very commonly used There are many different types of electric motors, however, due to their many advantages compared to other types of electric motors (simple structure, low cost, easy maintenance, reliable operation ), three – phase induction motor is the most commonly used The process of calculating and selecting electric motors for the drive system is done through the following steps of calculation:

- Engine power

- Number of preliminary synchronous revolutions of the motor

- Requirements on starting torque, overload and installation method

1.1 The engine power

- The engine power is calculated by this equation:

P ct = P

η t

+) Pct is the necessary engine power

+) Pt is the calculated power in the working

process +) η is the drive system efficiency

This is The necessary engine power (P ct ) equals to The calculated power in the working process (P t ) devided by The drive system efficiency (η).

The calculated power in the working process (P t ) is calculated as shown

below:

P t =P lv = F v

= 1800.2,47

=4,45(kW )

10001000

As F is The conveyor belt pulling force, v is The speed of conveyor belt From the Table 2.3[1], p.19, select The drive system efficiency (η) = 0.84

Calculating The necessary engine power (P ct ) as followed:

P ct = P

η t = 4,45

0.84 =5,30 (kW )

1.2 An overview of motor shaft revolution

- The revolution of final drive shaft:

n lv = 60000 v = 60000.2,47 =112(rev / min)

As The revolution of final drive shaft (n lv ) equals to 6000 times The velocity

of large sprocket (v) devided by Number of Large Sprocket Teeth (z) times

Sprocket pitch (p).

- From the Table 2.4[1], select:

+) Speed ratio of chain drive: =uxt=3

+) Gear ratio: ubr=4

- The whole drive transmission (u): u=uxt.ubr=3.4=12

- The revolution of motor shaft:

nem=u.nlv=112.12=1344 (rev/min)

Searching the Table P1.1[1], for a combination of 1344 rev/min and 5,3

kW provides a 4A112M4Y3 electric motor.

Results: The 4A112M4Y3 electric motor.

P = 5,5 kW, n = 1425 rev/min

1.3 Corrected transmission

u corrected =n em

= 1425

=12,7

n lv 112

The chain drive transmission (u xt ) is selected uxt =

3 The gear ratio (u br ) u br = u

=12,7

=4,2

u xt 3

All the torques (T), power transmissions (P), revolutions (n), drive transmission (u)

of each shaft is followed the Table 1.1 below:

Table1.1 Torques (T), power transmissions (P), revolutions (n), drive transmission

(u) of each shaft

P (kW) 5,24 5,14 4,94 4,45

n (v/p) 1425 1425 339 113

T (Nmm) 35117,19 34447,02 139165,2 376084,10

2 DESIGNING THE CHAIN DRIVE

When selecting roller chains, the following 7 parameters should be taken into

account

6

3.Prime Motor Type

Based on these conditions, Roller chain is selected.

2.1 Sprocket teeth

Using the selection guide table(Table 3)or the power transmission efficiency tables, select the chain and the number of small sprocket teeth that satisfy the rotary speed of the high-speed shaft and the corrected power transmission(kW) The chain pitch should be as small as possible, as long as the required power transmission efficiency is achieved This should minimize noise and ensure smooth transmission of power (If a single chain does not provide the required power transmission efficiency, use multiple chains instead If the installation space requires that the inter-shaft distance as well as the outer diameter of sprocket be minimized, use small-pitch multiple chains.) There should be a minimum wrap angle of 120˚ between the small sprocket and the chain

Based on these parameters and from Table 5.4[1]:

-Number of small procket teeth: z1 = 25

-Number of large procket teeth: z2 = 75

2.2 Main parameters of the chain

drive a) Pitch

From the Table 5.5[1] with these conditions from Table 1.1 provides the chain:

Pitch p = 25,4 mm

Roller diameter d c = 7,95 mm

Roller width B = 22,61 mm

Power transmission efficiency [P] = 19 kW

b) Distance between Shaft centers

a= p [x− z

1 + z

2 + √ (x− z

1 + z

2 )2 −2(z

2−z

1 )2 ]=1121,42 (mm)

2.3 All parameters of the chain drive

After calculating others parameters of the chain drive and looking up from

some tables All parameters of the chain drive are show in the table below

Table 2.1 All parameters of the chain drive

Distance between Shaft centers a 1122

Number of small sprocket teeth z1 25

Number of large sprocket teeth z2 75

Lực tác dụng lên trục Fr 1582,4 N

3 DESIGNING THE HELICAL GEARBOX

3.1 Materials

Based on Table 6.1[1], the material of each gear is selected below:

- The driving gear

+) Steel grade: C45

+) Heat treatment: Hardened after machining

+) Brinell hardness: HB = 241 ÷ 285, select HB1 = 265

+) Tensile strength: 850 MPa

+) Yield strength: 580 Mpa

-The driven gear +) Steel grade: C45

+) Heat treatment: Hardened after machining

+) Brinell hardness: HB = 192 ÷ 240, select HB2 = 230

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+) Tensile strength: 750 MPa

+) Yield strength: 450 Mpa

3.2 Calculating Contact stress [σ H ]and Bending stress [σ F ]

Contact stress [σ H 1 ]and Bending stress [σ F 1 ]of the driving gear

[σ H1]= σ

0

Hlim

Z R Z v K xH K HL =545,45( MPa)

S H

σ Flim0

[σ F1]= Y R Y S K xF K FL =272,57( MPa)

S F

Contact stress [σ H 2 ]and Bending stress [σ F 2 ]of the driven gear

[σ H2]=

σ0Hlim

Z R Z v K xH K HL =481,81( MPa)

S

H

σ Flim0

Y R Y S K xF K FL =236,57 ( MPa)

[σ F2]= S F

[σ H ]sb

Initial Contact stress

[σ H ]sb= [σ H1]+[σ H2 ]

= 545,45+ 481,81=513,63( MPa)

3.3 Center distance

3 T 1K

Hβ

a w =K a (u+1) √

[σ H]sb2uψ ba =106,07(mm)

So a w = 110 mm is chosen

3.4 Other parameters of gears

a) Module

m = (0,01 0,02).aW = (0,01 0,02).110= 1,1 2,2 (mm) From the table of standard

gear module, provides m = 2 (mm) b) Number of teeth

Assume β=10 °=¿ cos β=0,9848.

So the number of teeth of driving gear is

Z1 =2 a w

cos β

= 2.110 0,9848 =20,83

Z1 = 20 (teeth) is chosen

Z 2 = u.Z 1 = 4,2.20 = 84

Z2 = 84 (teeth) is chosen

c) Helix angle

cos β= m (Z 1+Z2)

=> β=arccos (cos β )=arccos(0,9636 )=19 °

d) Check for factor of safety

All the calculated strengths is less than allowable strengths, then the design is safe

e) Table of parameters of gears Table 3.1 Parameters of gears

Distance center a w 110 mm

Number of teeth Z 1 20

Pitch circle diameter d w1 42,3 mm

d w2 177,7 mm Tip diameter d a1 46,3 mm

d a2 181,7 mm Root diameter d f1 37,3 mm

d f2 172,68 mm

Contact force

Tangential force F t 1628,7 N

Radial force F r 626,97 N

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4 DESIGNING THE SHAFT

4.1 Initial caculations

a) Materials

- Symbol: C45

- Tensile strength: 600 Mpa

- Allowed torque strength: [τ] = 12 ÷ 30 Mpaτ] = 12 ÷ 30 Mpa] = 12 ÷ 30 Mpa

b) Distance between each point of force application

Fig 1: Distance between point of force application

l m 13=(1,2 ÷1,5 )d1=(1,2 ÷ 1,5)25=30 ÷37,5 (mm)

=> lm13 = 37 (mm)

l m 23=(1,2 ÷1,5 )d2=(1,2 ÷ 1,5)30=36 ÷ 45( mm)

=> lm23 = 40 (mm)

l m 12=(1,4 ÷ 2,5) d1=(1,4 ÷ 2,5) 25=35 ÷ 62,5(mm)

=> lm12 = 45(mm)

l m 22=(1,2÷ 1,5) d2=(1,2 ÷ 1,5)30=36 ÷ 45( mm)

=> lm22= 45 (mm)

l cki=0,5 (l mki +b0 )+ k3 +h n

¿

- Second shaft:

l 22=lc 22=70(mm)

l 23=0,5(l ¿¿ m 23+b02 )+ k1 +k2 ¿

Chọnl23=55( mm)

l 21=2 l23 =110(mm)

-First shaft:

l 12=lc 12=69(mm)

l 11=2.l13=110 (mm)

4.2 Diameter of the first shaft (I) caculation and

selection a) Internal torque diagram of the first shaft (I)

Fk14 = 153,1 (N)

Ft13 = 1628,7 (N)

Fr13 = 662,59 (N)

Fa13 = 560,81 (N)

F x 10=910,39 (N)

F y10 =223,62( N)

F x 11 =565,21( N )

F y11 =438,97( N )

Fig 2 Internal torque diagram of the first shaft (I)

b) Diameter of shaft at each point of force application

M j11=√ M2x 11+ M2y11=√10563,92 +02=10563,9( Nmm)

M tđ 11 =√M2j 11 +0,75 T211 =√10563,92+ 34447,022 =36030,45( Nmm)

M

tđ 11

= 3 36030,45=18,88 (mm)

M J 13 =√M 2x 13+ M 2y13=√50071,452 +24143,352=55588,23 (Nmm)

M tđ 13=√ M2J 13 +0,75 T213=√55588,232 +0,75.34447,022=63087,24 ( Nmm)

⇒d13=

3 M tđ13

=

3 55484,79

=21,55(mm)

√0,1 [σ] √ 0,1.63

M j 14=√M 2x14 + M2y 14=0 ( Nmm)

M tđ14=√M2j14 +0,75 T142=√0+0,75 34447,022=29831,99( Nmm)

⇒d14 = 3 M tđ12 = 3 29831,99=16,79( mm)

√0,1[σ] √ 0,1.63

d 10=d11 =18,8(mm)

Based on those calculations above, corrected diameter of shaft (I) at each point

of force application:

d 13=22( mm)

c) Key selection for the first shaft (I) From Table 9.1a[1] :

- At d13=22 mm, the parameters of key are:

b=6 mm

h=6 mm

t1=3,5 mm l=36 mm

- At d14=18 mm, the parameters of key are:

b=6 mm

h=6 mm

t1=3,5 mm l=36 mm

d) Check for factor of safety

All t he calculated actual factor of safety is greater than minimum factor of

safety, so the design is safe

4.3 Diameter of the second shaft (II) caculation and selection

Because of limited time, only the first shaft is calculated

Ft23 = 1628,7 (N)

Fa23 = 560,8 (N)

Fr23 = 626,97 (N)

F x20 =2728,53( N)

F y 20 =2109,04( N )

F x21=310,86( N ) F y 21=−731,16( N )

Fig 3 Internal torque diagram of the second shaft (II)

d20=d21=30 (mm )

d 23=32( mm)

- Key selection for the second shaft (II)

- At d13=32 mm, the parameters of key are:

b=10 mm

h=8 mm

t 1=5 mm l=36 mm

5 SELECTING AND ANALYZING THE

BEARINGS

5.1 Bearing selection

At position 10:

F r 10=F r 0 =√ F2x 10 + F2y10 =√910,392+ 223,622 =937,45( N )

At position 11:

F r 11 =F r 1=√F2x 11 + F2y11=√565,212+ 438,972=715,65( N )

Axial force:

16

F a =F a 13=560,81( N)

Because the axial force is huge, so Tapered roller bearing is chosen.

5.2 Bearing parameters for the first shaft (I)

From Table P2.11[1], with d = 20 mm:

Symbol: 7204

Inner diameter: d= 20 mm

Outer diameter: D= 47 mm

Dynamic load rating: C= 19,1kN

Static load rating: C0= 13,30kN

Cone witdth: B= 14mm

Cone angle α = 13,500

5.3 Analyzing the bearings

a) Dynamic load

C d =Q m√L=2,3361.10 /√3 855=17,7 ( kN )<C=19,1(kN )

b) Static load

Q t =1,17859<C 0=13,3(kN )

So all of those are allowed

Symbol: 7206

Inner diameter: d= 30 mm

Outer diameter: D= 62 mm

Dynamic load rating: C= 29,8kN

Static load rating: C0= 22,3kN

Cone width: B= 16mm

Cone angle α = 13,670

6 THE OUTSIDE STRUCTURES OF

GEARBOX

All of those parameter are shown in this drawing below

Duong Tien Luan_20184997_BV.pd

7 CONCLUSION

The academic goals of this thesis were initially uncertain and certainly change throughout the course of the year The initial aim was to just design (theorical way) transmission systems The goals and expectations for this thesis have been

achieved and it is hoped that the presented body of work allows for several new thesis topics to be researched in the future

8 REFERENCES

[τ] = 12 ÷ 30 Mpa1] Tính toán thiết kế hệ dẫn động cơ khí – Tập 1 – Trịnh Chất, Lê Văn Uyển – Nxh Giáo dục

[τ] = 12 ÷ 30 Mpa2] Tính toán thiết kế hệ dẫn động cơ khí – Tập 2 – Trịnh Chất, Lê Văn Uyển – Nxh Giáo dục

[τ] = 12 ÷ 30 Mpa3] Cơ sở thiết kế máy và chi tiết máy – Trịnh Chất – Nxb Khoa học và Kỹ thuật [τ] = 12 ÷ 30 Mpa4] Hướng dẫn làm bài tập dung sai – Ninh Đức Tốn, Đỗ Trọng Hùng – Trường ĐHBKHN

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