MEE 038: Machine Design 11

Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Lesson title: BELTING-FLAT (Open-Belting, Belt Materials) Materials:

Lesson Objectives: Module, Paper, Pen, Calculator.
References:
1. The major objective is to learn to apply basic theories of 1. Design of Machine Elements,
strength of materials and other pertinent theories to the 4e, by: Virgil Moring Faires
actual design of the most common machine elements; 2. Design of Machine Members,
Flexible Power-Transmitting Elements, Power Chains, Wire 4e
Ropes, Flywheel, Brakes and Clutches, etc. especially as By: Daughtie and Vallance
they are affected by the variation in the loading. 3.Required: Problem Book on
2. To give the safe answer hence by nature most machines, Design of Machine Elements 4e
the loads vary, and the most important consideration is to by: Virgil Moring Faires.
have minimum weight but do not compromise the 4. Machine Design Material and
durability, economy, and esthetics. Shop Practices. By SUBARAN.

“Focus on one goal at a time” ( by Steve Pogue)

A. LESSON PREVIEW/REVIEW
1) Introduction (2 mins)
As is true of the machine elements, Flexible connectors for transmitting power appear in
many different forms: flat belts, V-belts, flat-V, “toothed” belts, ropes (manila, cotton, wire).
The flexible drives have distinct advantages that are on occasion overwhelming: they
absorb vibration and shock, tending to transmit only a minimum to the connected shaft; they
suitable for relatively large center distances; they are quite; when properly maintained, they can
be designed to have a long trouble-free life.

2) Activity 1: What I Know Chart, part 1 (3 mins)

Before we start the topic answer the questions in column 2. Write your answers in the column 1. Leave
the third column blank (will be answered during Activity 4).

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

What I Know Questions: What I Learned (Activity 4)

1. In Drive selection data,
name at least four (4) type
drive.
2. What are density values in
lb. per in³ and kg per m³ of
the following: a) Leather, b)
canvas, c) rubber, d) balata,
e) single run belt, f) double
run belt
3. What is the recommended
Factor of Safety on Leather
Belt?
4. What is the value of the
ultimate stress on Leather Belt?
5. In belting what is the limit of
Arc of contact in degrees?

6. In Cross Belting what is the

limiting width in inches?

7. In kW transmitted, the power

would decrease at what per
cent for every degree greater
than what degree also.

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

B.MAIN LESSON

O-C belting.xspf

1) Activity 2: Content Notes (40 mins)

FLAT BELTING

Open Belt

Arc of contact

𝐷−𝑑
𝜃 = 180° ± [2𝑠𝑖𝑛−1 ( )] 𝑑𝑒𝑔
2𝐶

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

𝐷−𝑑
𝜃 = 𝜋 ±[ ] 𝑟𝑎𝑑
𝐶

Belt Length

𝜋 (𝐷 − 𝑑)2
𝐵. 𝐿. = (𝐷 + 𝑑) + 2𝐶 + [ ]
2 4𝐶

Belt Materials

a) Leather [Oak-tanned chrome]

Chrome tanning is a method whereby chromium sulphate is used to preserve animal skins and prevent rot. This
process gives a soft and very strong leather with many benefits and some drawbacks.

b) Fabric & Canvass

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

A construction of conveyor belt made up of plies of cotton fabric stitched together. Stitched canvas
belts may be untreated, impregnated, or coated.

c) Rubber belts

• General-purpose belting: This is for general conveying or power transmission.

• Agricultural belting: This is for agricultural applications like silage transfer, farm
equipment belts, etc.
• Retail belts: They used used in checkout counters or inventory transfer for various
commercial purposes.
• Construction belting: They are used for materials such as roofing shingles or plywood.
They are heavy-duty belts designed for use on construction equipment.

d) Balata belts

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Balata belting is a highly durable, robust belt made from heavyweight cotton fabric, impregnated with a high
quality rubber compound. It is pre-stretched during manufacture to ensure minimum stretch during operation.

The power transmitted is measured by the difference in tension on the tight and slack sides of the belt.

𝐹1 −𝐹𝑐
= 𝑒 𝑓𝜃 eq. 16-4 p 381 Vallance
𝐹2 −𝐹𝑐

𝑒 𝑓𝜃 − 1
𝐹1 − 𝐹2 = (𝐹1 − 𝐹𝑐 ) 𝑒𝑞. 16 − 5 𝑝 382 𝑉𝑎𝑙𝑙𝑎𝑛𝑐𝑒
𝑒 𝑓𝜃

Where:

e = 2.718
𝜃 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑤𝑟𝑎𝑝, 𝑟𝑎𝑑
𝐹1 = 𝑡𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛
𝐹2 = 𝑠𝑙𝑎𝑐𝑘 𝑠𝑖𝑑𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛

Simplification of eq. 16-6 p 382 Vallance

𝐹1 𝐹1
= 𝑒 𝑓𝜃 it follows that 𝐹2 =
𝑒𝑓𝜃
𝐹2

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Initial tension should range from 200 to 240 psi for leather belts and from 10 to 12 lb. per ply per inch of

width for rubber belts.

Considering centrifugal force

12𝜌𝑏 t𝑉 2
𝐹𝑐 =
𝑔
Where
𝑙𝑏
𝜌 = 𝐵𝑒𝑙𝑡 𝑤𝑒𝑖𝑔ℎ𝑡,
𝑖𝑛3
𝑏 = 𝐵𝑒𝑙𝑡 𝑤𝑒𝑖𝑔ℎ𝑡
𝑡 = 𝐵𝑒𝑙𝑡 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
𝑓𝑡
𝑉 = 𝐵𝑒𝑙𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦,
𝑠
𝑚
𝑔 = 𝐵𝑒𝑙𝑡 9.807
𝑠𝑒𝑐 2

𝝆 𝑉𝐴𝐿𝑈𝐸𝑆
kg
0.035 for leather 968.625
𝑚3
kg
0.044 for canvass 1217.7 3
𝑚
kg
0.041 for rubber 1134.675 3
𝑚
kg
0.040 for balata 1107.00 3
𝑚
kg
0.042 for single run belt 1162.35
𝑚3
kg
0.045 for double run belt 1245.375 3
𝑚
𝐹𝑐 = 𝑓𝑐 𝑏𝑡

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

𝑒 𝑓𝜃 − 1
𝑓1− 𝑓2 = (𝑓1− 𝑓𝑐 ) [ 𝑓𝜃 ]
𝑒
𝐹𝑐 = 𝑀𝑉 2
Where
𝑘𝑔
M - mass per length of belt, 𝑚
𝑚
𝑉 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑠

𝑘𝑔 𝑚 2
(𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 3 ) (𝑉, 𝑠 )
𝑚
𝑓𝑐 = [ ] , 𝑀𝑃𝑎 𝑚𝑒𝑡𝑟𝑖𝑐
106

2𝜋𝑇𝑛
𝑃= 𝑚𝑒𝑡𝑟𝑖𝑐
60
(𝐹1 − 𝐹2 )(𝑉)
𝐻𝑝 = 𝐸𝑛𝑔𝑙𝑖𝑠ℎ
550
Or

(𝐹1 − 𝐹2 )(𝑉)
𝑃= 𝑚𝑒𝑡𝑟𝑖𝑐
103

IMPORTANT
a) F.S. = 10
b) Ult. Stress in leather belt is 28 MPa
c) Joint factor – table 16-4 p 385 Vallance
d) Arc of contact should never be less than 155°
e) Cross belting should not be more than 8 inches wide.
f) Decrease kW transmitted by 1% for every degree greater than 60°
g) Coef. Of friction for belts table 16-5 p 386

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

h) If f is not given assume a value of 0.40 – 0.50

𝑘𝑊 (𝑓1− 𝑓𝑐 )(𝑉) 𝑒 𝑓𝜃 − 1
= [ 𝑓𝜃 ]
𝑚𝑚2 1000 𝑒
Reference:

Mech’l Engg. Reviewer, Vol 1.

Machine Design Materials and

Shop Practices.

(BELT DRIVES)

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

This document is the property of PHINMA EDUCATION

 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

This document is the property of PHINMA EDUCATION

 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

This document is the property of PHINMA EDUCATION

 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

This document is the property of PHINMA EDUCATION

 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

This document is the property of PHINMA EDUCATION

 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

This document is the property of PHINMA EDUCATION

 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

This document is the property of PHINMA EDUCATION

 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Example-1

Two flat belt pulleys having a center to center distance of 137 cm have drive diameter of 72 cm and 38

cm. (a) Determine the length of the belt if both pulleys will rotate both directions. (b) Find the arc of

contact on the small and big pulley. (c) Find the arc of contact of small and big pulley.

Given:
𝐷1 = 36 𝑐𝑚 𝐷2 = 72 𝑐𝑚 𝐶 = 137 𝑐𝑚

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Solution:
𝜋 (𝐷2 − 𝐷1 )2
(𝑎) 𝐿. 𝐵 = (𝐷2 + 𝐷1 ) + 2(𝐶) + ( )
2 4𝐶
𝜋 (72 − 36)2 𝑐𝑚2
𝐿. 𝐵 = (26 + 72)𝑐𝑚 + 2(137)𝑐𝑚 + ( ) = 𝟒𝟒𝟔 𝒄𝒎
2 4(137𝑐𝑚)

𝐷2 − 𝐷1
(𝑏) 𝜃1 = 180 − 2𝑠𝑖𝑛−1 ( )
2𝐶
72 − 36
𝜃1 = 180 − 2𝑠𝑖𝑛−1 ( )
2(137)
𝜃1 = 𝟏𝟔𝟓𝒐

𝐷2 −𝐷1
𝜃2 = 180 + 2𝑠𝑖𝑛−1 ( 2𝐶
)
72−36
𝜃2 = 180 + 2𝑠𝑖𝑛−1 (2(137))

𝜃2 = 𝟏𝟗𝟔𝒐

(𝑐) 𝑆1 = 𝐴𝑟𝑐 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑜𝑓 𝑠𝑚𝑎𝑙𝑙 𝑝𝑢𝑙𝑙𝑒𝑦:

𝜋
𝑆1 = 𝑅1 𝜃1 = (18𝑐𝑚) (165𝑜 𝑥 ) 𝑆1 = 𝟓𝟏. 𝟖𝟒 𝐜𝐦
180𝑜

𝑆2 = 𝐴𝑟𝑐 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑜𝑓 𝑙𝑎𝑟𝑔𝑒 𝑝𝑢𝑙𝑙𝑒𝑦:

𝜋
𝑆2 = 𝑅2 𝜃2 = (36𝑐𝑚) (195𝑜 𝑥 180𝑜 ) 𝑆2 = 𝟏𝟐𝟐. 𝟓𝟐 𝐜𝐦

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Problem 1.

A 600 mm pulley running at 370 rpm drives another pulley of the same diameter by means of a 10

mm thick leather belt. The power transmitted is 23 kW. If the center distance between shaft is 3 m. find

the belt length of this leather flat belt.

Solution:

WITHOUT CENTRIFUGAL EFFECT:

This is an open belt.

C = 3m

D = 600 mm

d = 600 mm

Finding belt length for open belting

𝜋 (𝐷 − 𝑑)2
𝐵. 𝐿. = (𝐷 + 𝑑) + 2𝐶 + [ ]
2 4𝐶

But D = d

So.

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

𝜋
𝐵. 𝐿. = (2𝐷) + 2𝐶 + 0
2

𝐵. 𝐿. = 𝜋𝐷 + 2𝐶

Substitute values

𝐵. 𝐿. = 𝜋(600) + 2(3000) = 7885𝑚𝑚 𝑜𝑟 7.885𝑚 𝑏𝑒𝑙𝑡 𝑙𝑒𝑛𝑔𝑡ℎ

CONSIDERING THE CENTRIFUGAL EFFECT

Belt required

Rpm = 370

Thickness = 10mm

C = 3m

D = d = 0.6m

Kw = 23

Belt width = ?

Belt length =?
𝐵. 𝐿. = 𝜋(600) + 2(3000) = 7885𝑚𝑚 𝑜𝑟 7.885𝑚 𝑏𝑒𝑙𝑡 𝑙𝑒𝑛𝑔𝑡ℎ

If f is not given assume a value of 0.40 – 0.50.

𝜃 = 𝜋𝑟𝑎𝑑 = 3.1416

𝑒 𝑓𝜃 = 2.7180.45(3.1416) = 4.11

0.45 is taken an average coef. of friction 0.40 – 0.50

Solving for Velocity, v

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

𝜋(0.6𝑚)/𝑟𝑒𝑣(370)𝑟𝑒𝑣/𝑚𝑖𝑛 11.62𝑚
𝑣 = 𝜋𝐷𝑛 = 𝑠𝑒𝑐 =
60 𝑚𝑖𝑛 𝑠

𝑓𝑐 = (𝑑𝑒𝑛𝑠𝑖𝑡𝑦)(𝑉𝑒𝑙)2

𝑘𝑔
(968.625) (11.62𝑚/𝑠)2
𝑓𝑐 = 𝑚3
106
= 0.13𝑀𝑃𝑎

𝑓1 = 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
𝑈𝑙𝑡. 𝑆𝑡𝑟𝑒𝑠𝑠
𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 =
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦
28𝑀𝑃𝑎
𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 =
10
𝑓1 = 2.8𝑀𝑃𝑎

𝑘𝑊 11.62(2.8 − 0.13) 4.11 − 1

2
= [ ]
𝑚𝑚 1000 4.11

𝑘𝑊
= 0.0235
𝑚𝑚2
23
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑏𝑒𝑙𝑡 =
0.0235

𝐴 = 979.69 𝑚𝑚2

10𝑏 = 979.6

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Belt cross
-section b
𝑏 = 98𝑚𝑚
10 mm

Belt required

10mm x 100mm x 788.5mm Leather belt.

Problem 2.
Cast-iron pulleys, 100 mm and 500 mm in diameters are to transmit 2.24 kW at 1200 rpm of the
smaller pulley, through a single-ply oak-tanned leather belt. What is the reasonable ratio of net tension?
What width of belt is necessary? What is the resultant bearing force?

SOLUTION:

If f = coefficient of friction = 0.30 for oak-tanned leather belt on cast-iron pulleys.

𝐷−𝑑
Θ = minimum angle of contact = 3.1416 – ( 𝐶
)

Assume C = 3 m or 3000 mm
500−100
3.1416 − ( ) = 3 𝑟𝑎𝑑
3000
𝐹 −𝐹 𝑓 −𝑓
𝑒 𝑓𝜃 = 𝑒 (0.30)(3.00) = 2.46 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 = 𝐹1 −𝐹𝑐 = 𝑓1 −𝑓𝑐
2 𝑐 2 𝑐

𝜋 𝑥 100 𝑥 1200
V = belt speed = 60 000
= 6.28 𝑚⁄𝑠

𝑓1 = 𝑚𝑎𝑥. 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑟 𝑙𝑒𝑎𝑡ℎ𝑒𝑟 𝑏𝑒𝑙𝑡𝑠.

28
= 10 = 2.8 𝑀𝑝𝑎
𝑘𝐺 𝑚
(𝑑, 3 )(𝑉, 𝑠 )2
𝑓𝑐 = 𝑐𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑚
1 000 000

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

𝑘𝐺 𝑚
(968.625 )(6.28 𝑠 )2
𝑓𝑐 = 𝑚3 = 0.0382 𝑀𝑃𝑎
1 000 000

𝑓2 = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑏𝑒𝑙𝑡

𝐹1 = 𝑡𝑖𝑔ℎ𝑡 − 𝑠𝑖𝑑𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛, 𝑁 = 𝑓1 𝑥 𝐴 = 𝑓1 𝑥 𝑏 𝑥 𝑡

𝐹2 = 𝑠𝑙𝑎𝑐𝑘 − 𝑠𝑖𝑑𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛, 𝑁 = 𝑓2 𝑥 𝐴 = 𝑓2 𝑥 𝑏 𝑥 𝑡

𝐹𝑐 = 𝑐𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑡𝑒𝑛𝑠𝑖𝑜𝑛, 𝑁 = 𝑓𝑐 𝑥 𝐴 = 𝑓𝑐 𝑥 𝑏 𝑥 𝑡

(a) Ratio of net tension = 𝑒 𝑓𝜃 = 2.46

𝑚 𝑓𝜃 −1)
(𝑉, )( 𝑓1 −𝑓𝑐 )(𝑒
(b) 𝑘𝑊⁄ 2 𝑜𝑓 𝑏𝑒𝑙𝑡 = 𝑠 (1000)(𝑒 𝑓𝜃 )
𝑚𝑚
= 0.0103 𝑘𝑊⁄
𝑚𝑚2

2.24
𝐴 = 𝑏𝑒𝑙𝑡 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 = = 218 𝑚𝑚2
0.0103
T = belt thickness = 3.175 mm of single-ply leather belts.
218
𝑏 = 𝑏𝑒𝑙𝑡 𝑤𝑖𝑑𝑡ℎ = = 68.66 𝑚𝑚
3.175
USE b = 70 mm

9550 𝑥 103 𝑥 2.24

T = torque transmitted = 1200
= 17 827 N.mm

2𝑇 2(17 827)
𝐹1 − 𝐹2 = = = 357 𝑁
𝑑 100

𝐹1 = 𝑓1 𝑥 𝐴 = (2.8)(218) = 610 𝑁
𝐹𝑐 = 𝑓𝑐 𝑥 𝐴 = (0.0382)(218) = 8.3 𝑁

𝐹1 − 𝐹𝑐
= 2.46 𝐹2 = 253 𝑁
𝐹2 − 𝐹𝑐

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Resultant bearing force = 𝐹1 + 𝐹2 = 863 𝑁

DID YOU KNOW?

… that Newton while a student at age 22, invented differential

and integral calculus, discovered the law of universal
gravitation, formulated the three laws of motion, developed
the new theory of light in just 18 months and set a record of

2) Activity 3: Skill-building Activities (with answer key) (18 mins + 2 mins checking)

Check your answers against the Key to Corrections found at the end of this SAS. Write your
score on your paper

A ventilating fan running at 500 rpm is driven by 20 kW, 900 rpm induction motor. The pulley
mounted on the motor shaft is 250 mm. Center to center distance is 1,50 m.
Determine:
a) The angle of wrap
b) The thickness, width, and length of belt required.

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

3. Activity 4: What I Know Chart, part 2 (2 mins)

What I Know Questions: What I Learned (Activity 4)

3. In Drive selection data, 1. Standard V belt
name at least four (4) type 2. Leather flat belt
drive. 3. Rubber fabric flat belt
4. Standard Roller Chain

4. What are density values in ‘a) 0.035 - 968.625

lb. per in³ and kg per m³ of b) 0.044 - 1217.7
the following: a) Leather, b) c) 0.041 - 1134.675
canvas, c) rubber, d) balata, d) 0.040 - 1107.00
e) single run belt, f) double e) 0.042 - 1162.35
run belt f) 0.045 - 1245.375
3. What is the recommended F.S. = 10
Factor of Safety on Leather
Belt?
4. What is the value of the Ult. Stress in leather belt is 28
ultimate stress on Leather Belt? MPa

5. In belting what is the limit of Arc of contact should never

Arc of contact in degrees? be less than 155°

6. In Cross Belting what is the Cross belting should not be

limiting width in inches? more than 8 inches wide.

7. In kW transmitted, the power

would decrease at what per Decrease kW transmitted by
cent for every degree greater 1% for every degree greater
than what degree also. than 60°

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Use another sheet of paper if necessary.

4) Activity 5: Check for Understanding (5 mins)

Check your answers against the Key to Corrections found at the end of this SAS. Write your
score on your paper.

QUIZ!!!

Problem 1.
1
A leather belt 6 inches by in. thick running at 4000 ft/min connects 12 in. and 60 in. diameter pulleys.
4
The angle of contact are 270𝑜 and 240𝑜 for small and larger pulley respectively. Coefficient of friction of
large pulley is 0.4 on small pulley 0.3. If the allowable tension is 100 lb per in., determine the maximum
horsepower that can be transmitted without considering centrifugal force.

Problem 2.
1
A flat belt 6 inches wide and 3
in. thick and transmit 20 hp. The center distance is 8 ft. The driving pulley
is 6 inches in diameter and rotates at 2000 rpm such that the loose side of the belt is on top. The driven
𝑙𝑏
pulley is 18 inches in diameter, the belt material is 0.035 𝑖𝑛3 and the coefficient of friction is 0.30.
Determine the belt net tension.

C. LESSON WRAP-UP
5) Activity 6: Thinking about Learning (5 mins

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Three things you learned:

1.
2.
3.
Two things that you would like to learn more about:
1.
2.
One question you still have:
1.

Check for Understanding (5 mins)

LET ME HEAR FROM YOU…

In your opinion, why do studying “BELTING” important for us as engineers?
Write your opinion on the space provided below.

____________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________

TEACHER'S GUIDE…
The following questions were recommended to be used to evaluate the
understanding of the students about the lesson. It is recommended
that they answer it on a separate sheet of paper.

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

KEY TO CORRECTIONS:
Activity 3:

A ventilating fan running at 500 rpm is driven by 20 kW, 900 rpm induction motor. The pulley mounted on
the motor shaft is 250 mm. Center to center distance is 1,50 m.
Determine:
c) The angle of wrap
d) The thickness, width, and length of belt required.

Solution:

If the pulleys are of the same material, slippage will occur on the smaller pulley, the angle of contact
being smaller.

𝐹1 = 𝑡𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 = 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑏𝑒𝑙𝑡 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑒𝑙𝑡
= 𝑆𝑎𝑙𝑙 (𝑏 𝑥 𝑡)
𝐹2 = 𝑠𝑙𝑎𝑐𝑘 𝑠𝑖𝑑𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛

𝐹1
= 𝑒 𝑓𝜃 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜
𝐹2

𝐷+𝑑
a) 𝜃 = π ± 𝐶

Smaller pulley

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________
𝐷+𝑑
𝜃𝑠 = π - 𝐶

𝑁1 𝑑 = 𝑁2 𝐷
900(250)
𝐷= = 450 𝑚𝑚
500

(450−250)
𝜃𝑠 = 3.1416 – [ 1500
] = 3.00 𝑟𝑎𝑑

𝜃𝑠 = 172.360 =

b) 𝑒 𝑓𝜃 =?
f= 0.45
𝜃 = 3 𝑟𝑎𝑑
𝑒 𝑓𝜃 = 𝑒 (0.45)(3.00)
= 3.857

𝐹1
𝐹2
= 3.857

𝐹1 = 3.857𝐹2 ⟶ ①

Tight-side, 𝐹1

Slack-side, 𝐹2

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

Net force = 𝐹1 − 𝐹2

Torque = Force x Distance

=(𝐹1 − 𝐹2 )(𝑟)
𝑑
= (𝐹1 − 𝐹2 ) ( 2 )

2𝜋𝑇𝑛
𝑘𝑊 =
60000
20(60000)
𝑇=
2𝜋(900)

𝑇 = 212.2 𝑁. 𝑚

(212.2)(2)
𝐹1 − 𝐹2 = 0.25
= 1697.6 ⟶ ②

Simplifying ① and ②
3.857𝐹2 − 𝐹2 = 1697.6
𝐹2 = 594.18 𝑁
𝐹1 = 3.857(594.18)
= 2291.75 𝑁

𝐹1 = 𝑆𝑎𝑙𝑙 𝑥 𝐴
= 𝑆𝑎𝑙𝑙 𝑥 𝑏 𝑥 𝑡

28 𝑀𝑃𝑎 28
𝑆𝑎𝑙𝑙 = 𝐹.𝑆.
= 10 = 2.8 𝑀𝑃𝑎

𝐹 2291.75 𝑁
b x t = 𝑆 1 = 2.8 𝑥 106 𝑁
𝑎𝑙𝑙 ⁄𝑚2

𝐹 2291.75 𝑁
𝑏 𝑥 𝑡 = 𝑆 1 = 2.8 𝑥 106 𝑁
𝑎𝑙𝑙 ⁄𝑚2

𝑏 𝑥 𝑡 = 818 𝑥 10−4 𝑚2

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

= 818 𝑚𝑚2

Assume t = 10 mm
818 𝑚𝑚2
𝑏= 10 𝑚𝑚
= 81.8 𝑚𝑚 𝑠𝑎𝑦 82 𝑚𝑚

𝜋 (𝐷 − 𝑑)2
𝐵. 𝐿. = (𝐷 + 𝑑) + 2𝐶 + [ ]
2 4𝐶
𝜋 (450 − 250)2
𝐵. 𝐿. = (450 + 250) + 2(1500) + [ ]
2 4(1500)
= 4099 56 𝑚𝑚

KEY TO CORRECTIONS:
Activity 5:

Problem 1.
1
A leather belt 6 inches by 4 in. thick running at 4000 ft/min connects 12 in. and 60 in. diameter pulleys.
The angle of contact are 270𝑜 and 240𝑜 for small and larger pulley respectively. Coefficient of friction of
large pulley is 0.4 on small pulley 0.3. If the allowable tension is 100 lb per in., determine the maximum
horsepower that can be transmitted without considering centrifugal force.

GIVEN:
1 𝑙𝑏
b = 6 IN. t = 4 𝐼𝑁. V = 4000 ft/min 𝑇1 = 100 𝑖𝑛

𝐷1 = 12 𝑖𝑛. 𝜃1 = 240𝑜 𝑓 = 0.3

𝐷2 = 60 𝑖𝑛. 𝜃2 = 270𝑜 𝑓 = 0.4

SOLUTION:
𝐹1
= 𝑒 𝑓𝜃
𝐹2

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 MEE 038: Machine Design 11
Module #1, Teachers’ Guide

Name: _________________________________________________________________ Class number: _______

Section: ____________ Schedule: ________________________________________ Date: ________________

𝑙𝑏
100 𝑖𝑛. 0.3(270𝑜)(
𝜋
)
=𝑒 180𝑜
𝑇2
𝑙𝑏
𝑇2 = 24.32
𝑖𝑛.
𝑙𝑏 𝑓𝑡
(100 − 24.32) (6𝑖𝑛. )(4000)
𝑃= 𝑖𝑛 𝑚𝑖𝑛
𝑓𝑡 − 𝑙𝑏
33000 𝑚𝑖𝑛 − 𝐻𝑝

𝑃 = 𝟓𝟓. 𝟎𝟒 𝑯𝒑

Problem 2.
1
A flat belt 6 inches wide and 3
in. thick and transmit 20 hp. The center distance is 8 ft. The driving pulley
is 6 inches in diameter and rotates at 2000 rpm such that the loose side of the belt is on top. The driven
𝑙𝑏
pulley is 18 inches in diameter, the belt material is 0.035 and the coefficient of friction is 0.30.
𝑖𝑛3
Determine the belt net tension.

GIVEN:
1
b = 6 IN. t = 3 𝑖𝑛. P = 20 Hp CD= 8 𝑓𝑡
𝑙𝑏
𝐵𝑒𝑙𝑡 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = 0.035 𝑖𝑛3
. 𝐷1 = 6 𝑖𝑛. 𝑓 = 0.3

𝑁2 = 2000 𝑟𝑝𝑚 𝐷2 = 18 𝑖𝑛.

SOLUTION:

396000 𝑙𝑏 − 𝑖𝑛⁄𝑚𝑖𝑛
𝑃 = 20 𝐻𝑝 ( ) = 630.25 𝑙𝑏 − 𝑖𝑛.
𝐻𝑝

T = (𝐹1 − 𝐹2 )𝑟
639.25 𝑙𝑏 − 𝑖𝑛.
𝐹1 − 𝐹2 = = 𝟐𝟏𝟎. 𝟎𝟗 𝑙𝑏𝑠.
3𝑖𝑛.

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