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DEPARTMENT OF EDUCATION AND TRAINNING

HOCHIMINH CITY UNIVERSITY OF TECHNOLOGY AND EDUCATION

THEORY OF MACHINE AND MACHINE DESIGN

CHATER 5: CHAIN DRIVES

20-Mar-23

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Target

1. Understanding the basic of
chain drives

2. Understanding dynamic of
the chain drives

3. Design the chain drives

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Theoretical contents

I. Overview

II. Sequence of calculation and
design the chain drives

III. Exercise

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4 Overview 20/3/2023
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1. Structure

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2
4

4 1

1. Driving sprocket 2. Driven sprocket 3. Chain 4. Chain tensioner

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Overview

2. Geometrical parameters

 d1, Z1 pitch diameter and number of teeth on the sprocket 1

 d2, Z2 pitch diameter and number of teeth on the driven sprocket 2
 a: Centre distance

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Overview

3. Operating principle.

 Movement and load are transmitted from the driving sprocket to the
driven sprocket based on the principle of matching in chain drive.

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7 Overview 20/3/2023
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3. Operating principle.

 A chain tensioner is applied to increase the performance of the drive

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8 Overview

4. Classification

 Based on structure :
 Roller chain.

 Bush chain.
 Inverted tooth chain.

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4. Classification

 Based on lines of chain:
 1 line chain.
 Multiple lines chain.

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10 Overview 20/3/2023
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4. Classification

 Based on the load mode:

Chain drive Conveyor chain Lifting chain

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11 synchronized chain drive

Overview
5. Advantages and disadvantages


 Advantages:

Far center distance transmission (<8m).
Less tension adjustment
small force effected on shaft
No sliding
Smaller structure compared to belt drives
Long lifetime.
Transmit to multiple shaft at the same time

(synchronized function)

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Overview
5. Advantages and disadvantages

 Disadvantages :

 Cause noise due to impact
 Unstable instantaneous ratio.
 Complex in manufacturing,

assembly and maintenance .

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13 Overview 20/3/2023

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6. Application

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Overview

pc

7. Broken patterns

 Hinge Abrasion  increasing pitch Out of chain.
pc

 Broken roller

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Theoretical contents

I. Overview

II. Sequence of calculation and
design the chain drives

III. Exercise


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Sequence of calculation and design the chain drives

1. Initial parameters

 P1 (kW) Power on the driving sprocket
 n1 (rpm) Rotation speed of the driving sprocket
u Ratio

2. Determine parameters of chain drives

 Chain selection: => Roller chain
 Number of teeth on sprocket:

Z1  29  2u
Z2  uZ1
(Roller chain11  15 < Z < 100 120)

Z should be odd number

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Sequence of calculation and design the chain drives
2. Determine parameters of chain drives


 Calculation pitch of chain directly by the following equation :

pc  2.823 KT1  6003 KP1
Z1p0 K x Z1n1p0 K x

 Based on satisfied standard abrasion and searched in Table 5.4 ([P] and n01)

Pt  K Kz Kn P1  P
Kx

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Sequence of calculation and design the chain drives

2. Determine parameters of chain drives

 Based on n1 to determine n01 (the closest given range)

 Based on n01 and Pt , [P] is selected in Table 5.4

pc [P]

 n01 n01=50 n01=200 n01=400 …..
  pc
[P]

pc Pt<[P]


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Sequence of calculation and design the chain drives

2. Determine parameters of chain drives n1=350 rpm

Pitch of Pin Length of Pt=5.5 kW
chain diameter Bushing Allowable power [P], kW, rotation speed of driving sprocket
(mm) dc,(mm)
(mm) n01 ,(rpm)

50 200 400 600 800 1000 1200 1600

12,7 3,66 5,80 0,19 0,68 1,23 1,68 2,06 2,42 2,72 3,20
12,7 4,45 8,90
12,7 4,45 11,30 0,35 1,27 2,29 3,13 3,86 4,52 5,06 5,95
15,875 5,08 10,11
15,875 5,08 13,28 0,45 1,61 2,91 3,98 4,90 5,74 6,43 7,55
19,05 5,96 17,75
25,4 7,95 22,61 0,57 2,06 3,72 5,08 6,26 7,34 8,22 9,65
31,75 9,55 27,46
38,1 11,12 35,46 0,75 2,70 4,88 6,67 8,22 9,63 10,8 12,7
44,45 12,72 37,19
50,8 14,29 45,21 1,41 4,80 8,38 11,4 13,5 15,3 16,9 19,3

3,20 11,0 19,0 25,7 30,7 34,7 38,3 43,8


5,83 19,3 32,0 42,0 49,3 54,9 60,0 -

10,5 34,8 57,7 75,7 88,9 99,2 108 -

14,7 43,7 70,6 88,3 101 - - -

22,9 68,1 110 138 157 - - -

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Sequence of calculation and design the chain drives

2. Determine parameters of chain drives

 Kx is coefficient of uneven load distribution depend on lines of chain :

Lines of chain 12 3 4
Kx 1 1.7 2.5 3

 Kz coefficient of the teeth number on the driving sprocket :

K z  Z01  25
Z1 Z1

 Kn coefficient of the rotating speed of the driving sprocket:

Kn  n01
n1


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Sequence of calculation and design the chain drives
2. Determine parameters of chain drives

K = Kr KaK0KdcKbKlv

 Kr – Coefficient of Dynamic load:
+ Smooth performance Kr =1,
+ Impact load Kr =1,2÷1,5;
+ Strong impact load Kr =1,8.

 Ka – Coefficient of center distance

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Sequence of calculation and design the chain drives

2. Determine parameters of chain drives

 K0 – Coefficient of the drive position. The angle taken by the
center line and the horizontal line is smaller than 60o K0 = 1,25.

 Kdc – Coefficient of ability to adjust chain tension:
+ Kdc = 1 for center distance adjustment

+ Kdc = 1,1 for employing chain tensioner
+ Kdc =1,25 for chain drive can not adjust tension

 Kb Coefficient of lubrication conditions:
+ Kb = 0,8 for continuously lubricating;
+ Kb = 1; for dripping lubricating
+ Kb = 1,5 for periodic lubricating.

 Klv – Coefficient of working shift a day

+ Klv = 1 for 1 shift

+ Klv = 1,12 for 2 shifts

+ Klv = 1,45 for 3 shifts

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Sequence of calculation and design the chain drives

2. Determine parameters of chain drives

 Pitch circle diameter d1  pc d2  pc
sin    sin   
 Z1   Z2 

 Chain link x must be even number


𝑥 = 2𝑎 + 𝑍1 + 𝑍2 + 𝑍2 − 𝑍1 𝑝𝑐
𝑝𝑐 2 2𝜋 𝑎

 Centre distance

a = 0.25𝑝 [𝑥 − + 𝑥 − −8 ]

Note: d2 should be smaller than 600(mm), or it is limited by the specific conditions.
Then, lines of chain can be increased to reduce d2

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Sequence of calculation and design the chain drives

3. Validation of chain drive dynamic
a. Instantaneous velocity and ratio of chain drive

 Instantaneous velocity:

v2  v1 cos        
cos 
Z1 Z1 Z2 Z2

 Instantaneous ratio:

Z cos
2  const
u t  Z1 cos




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Sequence of calculation and design the chain drives
3. Validation of chain drive dynamic

b. Average velocity and ratio chain

 Average velocity:

v1  4 Z1 pc n1
6.10

v2  4 Z 2 pc n2
6.10

 Average ratio chain:

u  n1  Z2
n2 Z1

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Sequence of calculation and design the chain drives


3. Validation of chain drive dynamic

 Checking the number of impact per second.

 To limit the impact kinetic energy E, the number of impacts per second is
defined as:

i  Z1n1  [i] [i] search in Table 5.6
15x

Table 5.6 The allowable number of impact in second
Type of chain Pitch of chain pc, (mm)

12,7 15,875 19,05 25,4 31,75 38,1 44,45 50,8

Roller chain 40 30 25 20 16 14 12 10

Inverted tooth chain 60 50 40 25 20 - - -

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Sequence of calculation and design the chain drives

4. Calculating chain by resistance.

 Condition: The pressure generated on the contact surface of the pin and


hinge p must be less than the allowable pressure [p] of material wrapped
p  F  [p]
bushing.

A

 Because of difference of experimental condition.

[p]= [p0 ]
K

Allowable pressure[p0] search in Table 5.3.

Pitch of chain Allowable pressure in hinge of chain[po], (MPa) rotating speed of
pc (mm) driving sprocket n1, (rpm)

50 200 400 600 800 1000 1200 1600 2000

12,7 ÷ 15,875 35 31,5 28,5 26 24 22,5 21 18,5 16
–
19,05 ÷25,4 35 30 26 23,5 21 19 17,5 15 –
31,75 ÷ 38,1 35 29 –
44,45 ÷ 50,8 35 26 24 21 18,5 16,5 15 –
20-Mar-23
21 17,5 15 – – –

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Sequence of calculation and design the chain drives
5. Force Analysis acting on shaft

 Initial force F0

F0 = Kf qm a g (N)

 Kf : Coefficient of chain deflection
+ Kf =1 for vertical position of chain drive
+ Kf =3 for angle smaller 400
+ Kf =6 for horizontal position of chain drive

 qm : Mass of 1m chain (kg).
 a: Centre distance (m).
 Auxiliary Fv.

Fv = qm v2 (N)

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Sequence of calculation and design the chain drives
5. Force Analysis acting on shaft

 Tension force in each side
F2 = F0 + Fv (N)
F1 = F2+ Ft (N)
Ft= 2T1/d1 (N)


 Force acting on shaft Fr
Fr = K m Ft (N)

Km is dependent on the angle between the center line with the horizontal line
+ Km=1.15 for angle smaller 400
+ Km=1 for angle from 400 to 900.

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30 Problems

Q1: There is two lines of Roller chain. Transmission Power P1= 5,5 Kw and the rotating
speed of driving sprocket n1=380rpm. Teeth on driving sprocket and driven sprocket
are Z1=23 and Z2=63, respectively. The working condition coefficient K=1,8. Allowable
power of roller chain are given in Table 1.

a. Based on standard abrasion, determine pitch of chain pc of the chain drive

b. Determine pitch circle diameters

Table 1. Allowable power of roller chain

Pitch of Allowable power [P], kW, rotating speed of driving sprocket n01(rpm)
chain
p (mm) 50 200 400 600 800 1000 1200 1600

12,7 0,35 1,27 2,29 3,13 3,86 4,52 5,06 5,95

12,7 0,45 1,61 2,91 3,98 4,90 5,74 6,43 7,55


15,875 0,57 2,06 3,72 5,08 6,26 7,34 8,22 9,65

15,875 0,75 2,70 4,88 6,67 8,22 9,63 10,8 12,7

19,05 1,41 4,80 8,38 11,4 13,5 15,3 16,9 19,3

25,4 3,20 11,0 19,0 25,7 30,7 34,7 38,3 43,8

31,75 5,83 19,3 32,0 42,0 49,3 54,9 60,0 -

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31 Problems 20/3/2023
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Q1: Solve
a. Determine pc:

+ 2 lines of chain: => Kx=1,7

+ n1=380rpm => n01=400rpm

Kz  Z01  25  1, 09 KKK
Z1 23  Pt  P1 z n  6,7 kW
n01 400 Kx select pc=19,05mm
Kn    1,05
n1 380
n01=400rpm

Pt  [P]  8,38 kW


b. Determine the pitch circle diameters :
𝑝

𝑑 = 𝜋 = 𝟏𝟑𝟗, 𝟗𝑚𝑚
sin(𝑍 )
𝑝

𝑑 = 𝜋 = 𝟑𝟖𝟐, 𝟐𝑚𝑚
sin(𝑍 )

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Bài tập ứng dụng
Q. 2

Roller chain drive has 2 lines of chain, pitch of chain pc=25,4; teeth on driving
sprocket Z1=23, Z2=63. Rotating speed of driving sprocket n1=380rpm.
Center distance a=40pc and coefficient K=1,7.
Determine:
1. Pitch circle diameters
2. Chain links
3. Maximum power P1 (kW)

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