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DEPARTMENT OF EDUCATION AND TRAINNING

HOCHIMINH CITY UNIVERSITY OF TECHNOLOGY
AND EDUCATION

THEORY OF MACHINE AND MACHINE DESIGN

Chapter 4: BELT DRIVES

14-Mar-23

2

Target

I. Overview of belt drives

II. Understand the mechanism of
transmission problems

III. Calculation and design
drives belts

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Theoretical contents

I. Overview

II. Sequence of calculation and
design the belt drives

III. Exercise

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Overview

1. Structure.

3

3
4

2
1

1. Driving pulley 2. Driven pulley 3. Belt 4. Idler pulley

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5 Overview

2. Operating principle.


 Depend on friction force between belt and pulley.
 Movement and energy are transmitted from driving pulley to driven pulley

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2. Operating principle.

 Belt tension generates friction force.

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3. Classification.

 Cross-section of belt:
• Flat belt.
• V belt.
• Ribbed belt.
• Round belt.
• Timing belt

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8 Overview

3. Classification.


 Flat belt :
• Cross section 𝑏𝑥𝛿 𝑚𝑚 .
• Include: leather belt, rubber belt, ...

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3. Classification.

 V belt :
• Isosceles trapezoid section, contact the belt groove with two sides.
• In structure, V-belt include:
 Top fabric 1
 Tension cords.
 Cushion rubber.
 Compression rubber 4.

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3. Classification.

 V belt :
• Cross-section is standardized.

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11 Overview

3. Classification.

 V belt:

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Overview

4. Advantages, disadvantages and range of use.

 Advantages:

• Working with high speed range.
• Simple structure.
• Smooth working.
• Against overload.
• Drive transmission for two far axes.

 Disadvantages:

• Large structural framework.
• Belt ratio is not stable.
• Large force acting on shaft.
• Low lifetime(1000h - 5000h).

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Theoretical contents

I. Overview

II. Sequence of calculation and
design the belt drives

III. Exercise

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Sequence of calculation and design the belt drives
1. Initial parameters

 P (kW) Power of driving shaft

 n (rpm) Rotation speed of driving shaft

u Ratio

2. Select belt and cross section.Rotation speed of faster shaft n, rpm the cross section of V belt
is selected by the Figure
5000 based on the given
parameters:
3150 + Power P

+ Rotating speed n
A

2000

1250 B
C

800

D

500

E

315

200 3.15 5 8 12.5 20 31.5 50 80 125 200 400
2

Power P, kW

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Sequence of calculation and design the belt drives
3. Parameters of the belt drive.


 d1, d2 (mm) : Pitch circle diameters

 a (mm) : Center distance

 α1, α2 : Angle of wrap

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Sequence of calculation and design the belt drives

4. Select diameter of pulley.

 Select diameter of
driving pulley 𝒅𝟏
according to Table
4.13

 Standardized to
select 𝑑 (mm): 63,
71, 80, 90, 100,
112, 125, 140, 160,
180, 200, 224, 250,
280, 315, 335, 400,
450, 500, 630, 710,
800, 900, 1000, ...

 Nên chọn 𝑑 ≈
1,2 𝑑 , chỉ khi

nào yêu cầu kích
thước thật nhỏ
gọn mới chọn
𝑑 =𝑑

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Sequence of calculation and design the belt drives
4. Select diameter of pulley.

a) Validation velocity of the belt

v1   .d1.n1 v1< 25 m/s for V - belts
60.1000 v1< 40 m/s for Narrow V belts

 Calculation and select pitch circle diameter of the driven pulley d2:

u  n1  d2  d2  d2  d1.u

n2 d1 1   d1

Chọn 𝑑 𝑡ℎ𝑒𝑜 𝑑ã𝑦 𝑡𝑖ê𝑢 𝑐ℎ𝑢ẩ𝑛 𝑔ầ𝑛 𝑣ớ𝑖 𝑔𝑖á 𝑡𝑟ị 𝑡í𝑛ℎ 𝑡𝑜á𝑛 𝑛ℎấ𝑡

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Sequence of calculation and design the belt drives

5. Length and center distance.

 Assumed Centre distance (a):

u 1 2 3 4 5 >6
0.95d2 0.9d2 0.85d2
a 1.5d2 1.2d2 d2

0,55d1  d2   h  a  2d1  d2 

Select center distance as based on table and can be obtained by the formula

 Length of belt (l):

Based on the assumed center distance as, the length of belt is obtained by

following Equation:  d1  d2  d2  d1 2

l  2as  2  4a s

Then, the belt length is standardized and given in Table 4.13

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Sequence of calculation and design the belt drives
6. Determine center distance

 Calculation exactly center distance (a):


Based on the selected length of belt, the center distance is re-defined by

where,   2  82 

a
4

  l   d1  d2 

2

 d2  d1 

2

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Sequence of calculation and design the belt drives
7. Angle of wrap.

𝛼 = 𝜋 − 2𝛽 𝑟𝑎𝑑
𝛼 = 𝜋 + 2𝛽 (𝑟𝑎𝑑)

 Condition: 𝛽 ≤ 10° ⇒ 𝛽 ≈ sin 𝛽 =

𝛼 = 𝜋− 𝑟𝑎𝑑 = 180° − . 57°


𝛼 =𝜋+ 𝑑 −𝑑 𝑟𝑎𝑑 = 180° + 𝑑 −𝑑 . 57°

𝑎 𝑎

 Validation of the wrap angle: 𝛼 ≥ 120° for the V belt

𝛼 ≥ 150° for the Flat belt

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Sequence of calculation and design the belt drives

8. Calculation the number of belt.

 Number of belt is calculated by formula: (≤ 6)

z  P1Kd
[P0 ]C ClCuCz

+ P1 (kW) Power of driving pulley 1  150...1800
+ P0 (kW) Power permit on Table 4.19 for V- belt

Table 4.20 for Flat belt
+ Kd: Coefficients of dynamic load Table 4.7
+ Cα: Coefficient of angle wrap’s effect Table 4.15
Or calculation according to formula: 1 0,0025(180  1) if
+ Cl: Coefficient of length’s effect Table 4.16


l: Real length of belt

l0: Experimental length of belt on Table 4.19 and 4.20
+ Cu: Coefficient of ratio’s effect Table 4.17
+ Cz: Coefficient of load’s effect Table 4.18

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Sequence of calculation and design the belt drives
8. Calculation the number of belt.

Table 4.15
α 180 170 160 150 140 130 120 110 100 90 80 70
Cα 1 0,98 0,95 0,92 0,89 0,86 0,82 0,78 0,73 0,68 0,62 0,56

Table 4.16
l/l0 0,5 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0 2,4
Cl 0,86 0,89 0,95 1,0 1,04 1,07 1,10 1,13 1,15 1,20

Table 4.17

u 1 1,2 1,6 1,8 2,2 2,4 ≥3
1,07 1,11 1,12 1,13 1,135 1,14
Cu 1


Table 4.18

z 1 2-3 4-5 6

Cz 1 0,95 0,9 0,85

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Sequence of calculation and design the belt drives
8. Calculation the number of belt.

Table 4.19 Value allowable power [P0] of V belt

Cross section Diameter of Velocity (m/s)

and laboratory driving pulley 3 5 10 15 20 25

length lo (mm) d1 (mm)

O 63 0,33 0,49 0,83 1,04 1,14
lo = 1320
90 0,46 0,64 1,17 1,54 1,8 1,88

112 0,48 0,75 1,33 1,78 2,12 2,3

112 0,7 1,08 1,85 2,4 2,73 2,85


A 125 0,78 1,17 2,0 2,75,2,9 3,08 3,26
lo = 1700
140 0,8 1,25 2,20 2,3,14 3,44 3,75

160 0,84 1,32 2,34 3,27 3,78 4,09

180 0,88 1,38 2,47 4,06 4,46

125 0,92 1,38 2,25 2,61 - -

lo = 2240 180 1,2 2,13 3,38 4,61 5,34 5,93

224 1,35 2,30 4,0 5,53 6,46 7,08

280 1,65 2,51 4,47 5,57 7,38 8,22

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Sequence of calculation and design the belt drives
8. Calculation the number of belt.

Table 4.19 Value allowable power [P0] of V belt

Cross section Diameter of Velocity (m/s)
and laboratory
length lo (mm) driving pulley 3 5 10 15 20 25

B d1 (mm) 6,28 -

lo = 3750 9,23 9,69
200 1,83 2,73 4,55 5,75 10,27 11
Г 11,33 12,27
lo = 6000 250 2,3 3,54 6,02 8 12,42 13,63
14,15 15,62
280 2,46 3,77 6,59 8,82
16,5 17,51
315 2,63 3,88 7,39 9,71 24,9 26,47
27,89 32,19
355 2,84 4,29 7,57 10,51 31,11 34,23

450 3,08 4,74 8,54 11,53

355 - 6,67 11,17 14,91

500 - 9,75 15,57 20,23

630 - 10,76 17,46 23,60

800 - 11,14 19,16 26,50

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Sequence of calculation and design the belt drives
8. Calculation the number of belt.

Table 4.20 Value allowable power [P0] of V belt


Cross section Diameter of Velocity (m/s)

and laboratory driving pulley 5 10 15 20 25
length lo (mm) d1 (mm) 3

63 0,71 0,93 1,46 1,77 1,85 -

71 0,77 1,15 1,85 2,46 2,72 2,69

YO 90 0,93 1,46 2,74 3,74 4,23 4,52
lo = 1600
112 1,15 1,73 3,15 4,26 5,23 5,85

140 1,46 1,88 3,54 4,93 6,14 7

180 2,23 4 5,74 6,87 7,28

180 2 3,05 5,33 7,53 9,15 10,26

YA 220 2,12 3,14 5,77 7,93 9,77 11,15
lo = 2500
224 2,23 3,26 6,02 8,46 10,3 11,85

250 2,34 3,72 6,61 8,77 10,85 12,55

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Sequence of calculation and design the belt drives

9. Sliding phenomenon of belt drive.

 Sliding curve ε(ψ) – effect curve η(ψ).
Set coefficient of tension:   Ft   t
2F0 2. 0

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Sequence of calculation and design the belt drives
9. Sliding phenomenon of belt drive.

 Elastic Sliding :

Cause of the belt is
elastic strain. The more
flaccid of belt, the more
stretched so sliding
increase

 Slippery Sliding : AC. Sliding arc BC. Static arc

Cause of the belts chains is overload, tangential force Ft is higher than friction Fms,
The sliding slippery occurred at ACB arc on pulley

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Sequence of calculation and design the belt drives
10. Tension in the belt

 Tension in the belt:

Or Fr = 2.F0.sin(α1/2) 14-Mar-23
(Common Fr = ( 2 – 3)Ft

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Sequence of calculation and design the belt drives
10. Tension in the belt

• Tangential force cause by T1:

• Tension of the side F1: F1 = F0 + Ft /2.
• Tension of the side F2: F2 = F0 – Ft /2.
According to Euler’s formula about friction,
We have: F1 = F2.efα .

Ft (e fα 1) Ft  2
F0   fα   1  fα 
2 (e 1) 2  (e 1) 

F0 Initial force is necessary for transmitting power P1 or torqueT1.

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Sequence of calculation and design the belt drives
11. Stresses in belt.

 Bending stress:

If belt material according Hook’s law: σu = E.ε

(E: Young module; ε Elongation)

 Stress diagram: δ
ε=

𝑑

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Sequence of calculation and design the belt drives
11. Stresses in belt.

 Tensile stress:

 Stress due to initial tension:  0  F0
 Stress tension on driving side: A
 Stress tension on driven side:
 Stress due to centrifugal forces: 1  F1   0   t ; t  Ft
A 2 A


 2  F2   0  t
A 2

FV qm.v2 (condition v>20m/s)
V  

AA

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Exercise

 Q.1
There is the flat belt drives with Power P1=6,5kW and the rotation speed of
the faster shaft n1=1420 rpm and the remained shaft n2=565 rpm. The
center distance a is 1800mm. The velocity of belt is from 14,5 m/s to 15,5
m/s. The pitch circle diameter of pulleys are standardized in the given
range: 160, 180, 200, 224, 250, 280, 315, 400, 450, 500, 560, 630, 710
(mm).
1) Determine diameter of driving pulley d1 and driven pulley d2?
2) Validate the wrap angle α1?

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Exercise


1) Determine diameter of driving pulley d1:

Solution 1: Based on velocity of belt

vmin  πd1minn1  d1min  60000vmin  60000 14, 5  195,12mm
60000 πn1 3,14 1420

vmax  πd1maxn1  d1max  60000vmax  60000 15, 5  208,57mm
60000 πn1 3,14 1420

Select d1 based on standard values: d1 = 200mm

Solution 2: Based on Savơrin formula

d1  (1100 1300) 3 P1  182, 64  215,85(mm)
n1

Select d1 based on standard values : d1 = 200mm

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Exercise

1) Determine diameter of driven pulley d2:

u12  n1  d2  d2  n1 d1  502, 65mm
n2 d1 n2


Select d2 base on standard values : d2 = 500mm

2) Checking the condition of wrap angle α1:

α1  1800  d2  d1 570  1800  500  200 570  170,50  1500
a 1800

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Exercise

 Q2:
The flat belt drive includes the pulleys with pitch circle diameters
d1=150mm, d2=600mm. Width belt b=50mm and thickness = 6mm. The
friction coefficient between the pulley and belt is 0,32; Initial stress
0=2MPa. Angle of wrap 1=1600 and rotating speed n1=1000v/ph.

 Determine center distance a of belt drives?
 Determine initial tension F0 and tangential force Ft on driving pulley?
 Calculating max power P1?

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Exercise


 Determine center distance a of the belt drives :

𝑑 −𝑑
𝛼 = 180 − 57

𝑎

𝑎 = 57 𝑑 −𝑑 = 57 600 − 150 = 𝟏𝟐𝟖𝟐, 𝟓𝑚𝑚

180 − 160 20

 Determine initial tension 𝐹 and tangential force 𝐹 on driving pulley :

• Cross-section: 𝐴 = 𝑏 ∗ 𝛿 = 50 ∗ 6 = 𝟑𝟎𝟎𝑚𝑚
• 𝐹 =A0= 300x2=600 N

 tangential force Ft :

• Condition tension belt 2𝐹 : 𝑒 − 1 ≥ 𝐹 𝑒 + 1

2𝐹 𝑒 − 1 2 ∗ 600 ∗ 𝑒 , ∗ , − 1
𝐹= = = 𝟓𝟎𝟑, 𝟏𝑁
+1 𝑒 , ∗ , +1
𝑒

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Exercise


 Calculating max power P1 :
• Velocity on driving pulley:

𝑣= 𝑑 𝑛 3.14 ∗ 150 ∗ 1000
= = 𝟕, 𝟖𝟓(m/s)
60. 10 60. 10

• Power on the driving pulley:

𝑃= 𝐹 𝑣 503,1 ∗ 7,85
= = 𝟑, 𝟗𝟓𝑘𝑊
1000 1000

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Exercise

 Q3:
Flat belt drives has center distance a=1800mm, pulley diameter
d1=200mm, d2=600mm. Friction coefficient between pulley and belt
f=0,30; Power P1=6,5kW and rotating speed n1=1200rpm.

 Determine and checking the condition of wrap angle α1?
 Determine initial tension F0 to the belt chains disappear sliding

slippery
 Replace the belt by the other type with the fiction coefficient f=0,35.


How the tangential force Ft is increasing?

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Exercise

1) Determine and validation the condition of wrap angle α1

α1  1800  d2  d1 570  1800  600  200 570  167,30  1500
a 1800

2) Determine initial tension F0 to satisfy the working basic condition

α1=167,30 = 2,92 rad

Ft  2T1  517,3N
d1

where, T1  9,55106 P1  9,55 106 6,5  51730 Nmm
n1 1200

Hence,

F (e fα1  1) 517,3 (e0,32,92  1)
F0  t2(e fα1 1)  2  (e0,32,92 1)  627, 7N

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Exercise

3) How is the tangential force Ft change?

' 2F0 (e fα1 1) 2  627, 7  (e0,352,92  1)
Ft    591,04N
fα1 0,352,92  1)
(e 1) (e

The tangential force Ft increasing
Ft'  591,04  1,143
Ft 517,3

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