Chapter 4 Fundamental Of Component Design.pdf

Microsoft PowerPoint Chapter 4 FUNDAMENTAL OF COMPONENT DESIGN 15/9/2021 1 15 Sep 21 THEORY OF MACHINE AND MACHINE DESIGNChapter 4 FUNDAMENTAL OF COMPONENT DESIGN 1 DEPARTMENT OF EDUCATION AND TRAINNI[.]

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THEORY OF MACHINE AND MACHINE

DESIGN

Chapter 4:

FUNDAMENTAL OF COMPONENT DESIGN

DEPARTMENT OF EDUCATION AND TRAINNING

HOCHIMINH CITY UNIVERSITY OF TECHNOLOGY

AND EDUCATION

Contents

I Loads and stresses in machine parts

II Basic criteria of machine part

III Mechanical transmission system and their principal feature

2

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TS Phan Công Bình binhpc@hcmute.edu.vn binhpc.tpm@gmail.com 15-Sep-21

1 Loads

- Load in mechanisms is the physical unit on mechanical system or component

- Classification: Static load and dynamic load

TS Phan Công Bình binhpc@hcmute.edu.vn binhpc.tpm@gmail.com

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1 Loads

• 3 kinds of load:

 Rated load: The longest or maximum effective load

 Equivalent load: a constant value represented for dynamic load

 Assumed load: applied to design machine partstd dn E

Q Q K

tt td tt d dk

Q Q K K K

Ke: Coefficient depends on load mode

Ktt: Coefficient considers nonuniform distribution

Kd: Coefficient of dynamic load

Kdk: Coefficient depends on working condition

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2 Stress

- Stress is internal resistance due to external forces

Static stress Dynamic stress

2 Types of stress:

6

2 Stress

- Stress cycle: The minimum time for the stress to return to original state

- 2 types of cycle stress:

Cycle stress stable over time

Cycle stress oscillated over time

max min

2

a

  

max min

2

m

  

min max

r 





- Amplitude stress:

- Mean stress:

- Stress ratio:

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2 Stress

Symmetrical period asymmetrical period

opposite sign

asymmetrical period same sign Closed circuit period Constant stress

σ

σm

σa

σm

r = -1

r = 0 r = +C

r = -C

n

σm

e)

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2 Stress

( ) [ ]

k n

F

A

W

    [ ]

o

F W

   [ ]

F A

F: Force (N) and A: Cross-section (mm2)

W, Wo: The resistance moment (mm4)

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2 Stress

10

2 Stress

Contact stress

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Contents

III Mechanical drive transmission system 1

15-Sep-21

12

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1 Strength

Definition: The ability of material to resist breaking

2 types of destruction:

 Destruction caused by exceeding working stress (overload)

 Destruction caused by long-tern effects of changing stress value is

over the limit of ultimate strength

Calculate method: Static stress of element

 Plasticity material:

 Brittle material:

ε: Coefficient of measurement

Ks: Coefficient of concentration static load

[ ] ; [ ]



     

[ ] ; [ ]



     

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1 Strength

14

1 Strength

Element has dynamic stress σa, σm= const Calculation stress:

KL: Life coefficient depends on N

β: coefficient of durable increase

Calculate cycle of operations element:

 N > N0: Limited stress is limited long fatigue KL=1

[s]: 1.5-2 safety coefficient

 N < N0: Limited stress is limited short fatigue

lim

[ ] [ ] KL

s K

 

   

60 h LE k

N L n N N

[ ]

r

s K

 

   

0

[ ]

L L

LE

N

K K

 

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1 Strength

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1 Strength

Curved of fatigue

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2 Rigidity

Definition: The ability of material to resist deformation

Calculate:

 Axial force: Long transposed

E: Modulus of elasticity (MPa)

A: Axial cross-section

L: Length of axis

 Moment of deflection

y: displacement

θ: Rotate angle

 Moment of torsion

G: Modulus of elasticity

J0: Inertia moment

[ ]

a

F l

EA

   

[ ]; [ ]

y y  

0

[ ] Tl GJ

  

18

3 Wear strength

Definition: The ability of element that working under time of limited wear

Calculate:

 Limited pressure: Relationship between pressure and distance

 Condition: p ≤ [p]

m

P S const

4 Heating

During the working process, the machine will generate heat due to friction,

engine, …

The harmful effects of heat:

 Changing characteristic of material

 Machinery parts will stick together

 Reduce accuracy of machine because of thermal deformation

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4 Heating

II Basic criteria of operating capacity of machine part

Cutting temperature generation

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5 Resistance vibration

𝐹𝑙𝑡= 𝑐𝑦

𝑐 = 48𝐸𝐽

𝑙

where,

𝑒 𝑐

at 𝜔 = 𝜔𝑛= (natural frequency)

then 𝑦 → ∞ (𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒)

stiffness

Note: r𝐞𝐬𝐨𝐧𝐚𝐧𝐜𝐞 𝐟𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝐲𝝎𝝎

𝒏

≈ (𝟎 𝟕 ÷ 𝟏 𝟐)

l/2 l/2

F lt

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5 Resistance vibration

Contents

III Mechanical drive transmission system

22

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2 Velocity

• Velocity of driving pulley:

• Velocity of chain:

• Rotating speed of pulley:

• Rotating speed of chain:

• Angular velocity:



 ( / ) 60000

Dn

2 ( / ) 60

n rad s





60000v( )

D





60000v( )

zp

D: Diameter (mm) n: Rotating speed (rpm) p: Pitch of chain z: Number of teeth in sprockets

( / ) 60000

pzn

III Mechanical drive transmission

1 Power

( )

1000

t

Fv

P kW Ft: Tangential force (N)

v: Velocity (m/s)

v

F t

D

n

T

P

x

driven

P1

I

In Drive

T1

n1

T2

n2

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• Given Power and rotation speed

6 1 1

1

9.55 10

x xP

n



 1

2

t

d

3 Torque

4 Transmission Ratio

• Belt drives, friction drives:

2 12

1

(1 )

d

u

d







2

12

1

z

u

z



12 I II

n u n

• Given tangential Force and diameter

• Gear drives, worm-gear drives, chain drives:

12  II

I P P

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5 Efficiency

m

• Series:

• Parallel:

1 .2 3 n

TII= u ɳTI





12

II

I

P

1000 t Fv

6 0 0 0 0

v

D





n

26

ESSAY

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M

x

Plv (Ptđ)

Pct

x

II

15-Sep-21 15-Sep-21

 nt o,  br o,  x,o  unt  ubr  ux

ht  d1 12 23 uht  ud1 u12 u23

PII

PI III Mechanical drive transmission

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15-Sep-21

1 Determine the required power on the motor’s shaft

where, the reference efficiencies are shown on the given Table

η η   η  η 3 η

nt br ol x

Based on diagram, the system includes 1 clutch, 1 pair of helical gear, 3

bearing roller gear and 1 chain drive

η

 lv

ct

P

usb= uhx ux

2 Determine the preliminary speed of motor’s shaft

where, the reference ratios are shown on the given Table (minimum values)

nsb= usbx nlv

Based on diagram, the system includes 1 pair of helical gear and 1 chain drive

compared to 750; 1000; 1500 & 3000 (synchronous speed)

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3 Transmission Ratios

 m  

lv

n

br

u where u is chosen in the givenTable u

u



tt

u u

u compared to the required tolerance

u

Nếu chọn un theo tiêu chuẩn mà không thỏa sai số tỉ số truyền thì dùng luôn

giá trị đã tính được sau khi chọn uh

Try to select uxbased on the rated range

Check the tolerance of total ratio

Nên chọn un (udhoặc ux) theo tiêu chuẩn rồi kiểm tra sai số tỉ số truyền tổng thể

M

x

30

I

II

III

đc

Example

Giving the parameter on the working shaft:

T = 425000 Nmm, n = 120 rpm

1 Motor specification

2 Transmission ratio distributions

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Motor selection

6

9,55.10 425000 120 5,34( )

9,55 10 9,55 10

lv

n

The load is static: Pt= Plv= 5,34 kW

1 Determine the power of working shaft

As shown on Table 2.3

ηd= 0,96 belt drive efficiency

ηol= 0,99 bearing efficiency

ηbr= 0,97 1 pair spur efficiency

ηnt= 0.9 clutch efficiency

η η η η η  d .ol3 br nt

Based on diagram: 1 belt drive, 3 bearing roller, 1 pair of helical gear, 1 clutch

η

0,81

t

ct

P

0,96 0,99 30,97 0,9 0,81 

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15-Sep-21

Spur gear drive

Bevel gear drive

Worm gear drive

- Self – locking

- Unself – locking with

z1=1

z1 = 2

z1= 4

Chain drives

Friction drives

Belt drives

Pair of rolling bearing

Pair of plain bearing

0,96 – 0,98 0,95 – 0,97

0,3 – 0,4

0,7 – 0,75 0,75 – 0,82 0,87 – 0,92 0,95 – 0,96 0,9 – 0,96

0,99 – 0,995 0,98 – 0,99

0,93 – 0,95 0,92 – 0,94

0,2 – 0,3

0,9 – 0,93 0,7 – 0,88 0,95 – 0,96

Table 2.3

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4 Calculating the preliminary speed of motor

Preliminary ratio (based on Table 3.2) Gear Ratio

Spur gear

- 1 level gear box

- 2 level gear box Bevel gear box

- 1 speed gear box

- 2 speed gear box Flat belt drive

- Simple

- Tensioner

V belt drive Chain drive Friction belt drive

1,6– 8

8 – 40

1 – 6,3

8 – 30

2 – 5

4 – 6

2– 5

2 – 5

2 – 4

Motor selection

Selection of motor (3 phases 380V-50Hz)

is satisfied by:

Pđc≥ Pct= 6.59 kW

nđc≥ nsb(~ 750 rpm)

Selection of electric power in Catalogue

usb= uhx uđ

= 1.6 x2(select minimum value)

nsb= usb,.nlv,= 120x3.2 = 384 rpm

nsb< 750selection Motor~ 750 rpm

34

Motor selection

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1 Ratio

Distribution the transmission ratio

 720 6 

120

m

lv

n

2,5

tt

br

u Select u u

u

    6 (2,5 2,5)4,1% 4% not

6

u u u u u

𝑢đ= 2; 2,24;2,5; 2,8; 3,15; 3,56; 4; 4,5 & 5

Chọn udtheo tiêu chuẩn không thỏa thì thì phải thiết kế theo số liệu đã tính

ud=2,4

Try to select ud=2,5 (based on the given rate range)

Then, check the tolerance of total ratio

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15-Sep-21

2 Rotation speed of shafts

Rotation speed of shaft I:

Rotation speed of shaft II:

Rotation speed of working shaft:

1

720 300 2,4

dc d

n

u

 1  

2

300 120 2,5

br

n

u

 2120

lv

3 Calculating the power on shafts

Power on shaft II:

Power on shaft I:

Power on motor’s shaft :

η η

2

5,34

6,0 0,9 0,99

lv

P

x

η η

1

6 6,18 0,98 0,99

P

x

η η

 1  6,18 6,5 0,96 0,99

dc

P

x

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4 Calculation torque on shafts

Motor’s shaft:

Shaft I:

Shaft II:

Working shaft :

9,55 10 6 9,55 106 6,586215

720

dc dc

dc

P

n

1

9,55 10 9,55 10 6,18 196730

300

P

n

2

9,55 10 9,55 10 6,0

477500 120

P

n

9,55 10 6 9,55 106 5,34424975

120

lv lv

lv

P

n

38

5 Results

Shaft

Rotation speed (rpm) 720 300 120 120

Moment (N.mm) 86215 196730 477500 424975

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Tiêu đề	Fundamental Of Component Design
Tác giả	TS. Phan Công Bình
Trường học	Hochiminh City University Of Technology And Education
Chuyên ngành	Theory Of Machine And Machine Design
Thể loại	chapter
Năm xuất bản	2021
Thành phố	Hochiminh City

Định dạng
Số trang	20
Dung lượng	2,7 MB