This document provides solutions to problems involving belt drives. It first solves for the tensions and power transmission in a belt drive system connecting two pulleys of different diameters, one running at 200 rpm. Taking into account centrifugal tension, friction, and a maximum tension of 2 kN, it finds the transmitted power is 13.588 kW. It also calculates the efficiencies lost to friction in the system.
This document is about power transmission system. It's aimed those interested in learning about mechanical engineering and students who are studying various programmes in engineering. This document only deals with power transmission through flat and v-belts.
ME010 801 Design of Transmission Elements
(Common with AU010 801)
Teaching scheme Credits: 4
2 hours lecture, 2 hour tutorial and 1 hour drawing per week
Objectives
To provide basic design skill with regard to various transmission elements like clutches, brakes, bearings and
gears.
Module I (20 Hrs)
Clutches - friction clutches- design considerations-multiple disc clutches-cone clutch- centrifugal clutch -
Brakes- Block brake- band brake- band and block brake-internal expanding shoe brake.
Module II (17 Hrs)
Design of bearings - Types - Selection of a bearing type - bearing life - Rolling contact bearings - static
and dynamic load capacity - axial and radial loads - selection of bearings - dynamic equivalent load -
lubrication and lubricants - viscosity - Journal bearings - hydrodynamic theory - design considerations -
heat balance - bearing characteristic number - hydrostatic bearings.
Module III (19 Hrs)
Gears- classification- Gear nomenclature - Tooth profiles - Materials of gears - design of spur, helical,
bevel gears and worm & worm wheel - Law of gearing - virtual or formative number of teeth- gear tooth
failures- Beam strength - Lewis equation- Buckingham’s equation for dynamic load- wear loadendurance strength of tooth- surface durability- heat dissipation - lubrication of gears - Merits and
demerits of each type of gears.
Module IV (16 Hrs)
Design of Internal Combustion Engine parts- Piston, Cylinder, Connecting rod, Flywheel
Design recommendations for Forgings- castings and welded products- rolled sections- turned parts,
screw machined products- Parts produced on milling machines. Design for manufacturing - preparation
of working drawings - working drawings for manufacture of parts with complete specifications including
manufacturing details.
Note: Any one of the following data book is permitted for reference in the final University examination:
1. Machine Design Data hand book by K. Lingaiah, Suma Publishers, Bangalore/ Tata Mc Graw Hill
2. PSG Design Data, DPV Printers, Coimbatore.
Text Books
1. C.S,Sarma, Kamlesh Purohit, Design of Machine Elements Prentice Hall of India Ltd NewDelhi
2. V.B.Bhandari, Design of Machine Elements McGraw Hill Book Company
3. M. F. Spotts, T. E. Shoup, Design of Machine Elements, Pearson Education.
Reference Books
1. J. E. Shigley, Mechanical Engineering Design, McGraw Hill Book Company.
2. Juvinall R.C & Marshek K.M., Fundamentals of Machine Component Design, John Wiley
3. Doughtie V.L., & Vallance A.V., Design of Machine Elements, McGraw Hill Book Company.
4. Siegel, Maleev & Hartman, Mechanical Design of Machines, International Book Company
Unit 6- spur gears, Kinematics of machines of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
This document is about power transmission system. It's aimed those interested in learning about mechanical engineering and students who are studying various programmes in engineering. This document only deals with power transmission through flat and v-belts.
ME010 801 Design of Transmission Elements
(Common with AU010 801)
Teaching scheme Credits: 4
2 hours lecture, 2 hour tutorial and 1 hour drawing per week
Objectives
To provide basic design skill with regard to various transmission elements like clutches, brakes, bearings and
gears.
Module I (20 Hrs)
Clutches - friction clutches- design considerations-multiple disc clutches-cone clutch- centrifugal clutch -
Brakes- Block brake- band brake- band and block brake-internal expanding shoe brake.
Module II (17 Hrs)
Design of bearings - Types - Selection of a bearing type - bearing life - Rolling contact bearings - static
and dynamic load capacity - axial and radial loads - selection of bearings - dynamic equivalent load -
lubrication and lubricants - viscosity - Journal bearings - hydrodynamic theory - design considerations -
heat balance - bearing characteristic number - hydrostatic bearings.
Module III (19 Hrs)
Gears- classification- Gear nomenclature - Tooth profiles - Materials of gears - design of spur, helical,
bevel gears and worm & worm wheel - Law of gearing - virtual or formative number of teeth- gear tooth
failures- Beam strength - Lewis equation- Buckingham’s equation for dynamic load- wear loadendurance strength of tooth- surface durability- heat dissipation - lubrication of gears - Merits and
demerits of each type of gears.
Module IV (16 Hrs)
Design of Internal Combustion Engine parts- Piston, Cylinder, Connecting rod, Flywheel
Design recommendations for Forgings- castings and welded products- rolled sections- turned parts,
screw machined products- Parts produced on milling machines. Design for manufacturing - preparation
of working drawings - working drawings for manufacture of parts with complete specifications including
manufacturing details.
Note: Any one of the following data book is permitted for reference in the final University examination:
1. Machine Design Data hand book by K. Lingaiah, Suma Publishers, Bangalore/ Tata Mc Graw Hill
2. PSG Design Data, DPV Printers, Coimbatore.
Text Books
1. C.S,Sarma, Kamlesh Purohit, Design of Machine Elements Prentice Hall of India Ltd NewDelhi
2. V.B.Bhandari, Design of Machine Elements McGraw Hill Book Company
3. M. F. Spotts, T. E. Shoup, Design of Machine Elements, Pearson Education.
Reference Books
1. J. E. Shigley, Mechanical Engineering Design, McGraw Hill Book Company.
2. Juvinall R.C & Marshek K.M., Fundamentals of Machine Component Design, John Wiley
3. Doughtie V.L., & Vallance A.V., Design of Machine Elements, McGraw Hill Book Company.
4. Siegel, Maleev & Hartman, Mechanical Design of Machines, International Book Company
Unit 6- spur gears, Kinematics of machines of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Deflection of curved beam |Strength of Material LaboratorySaif al-din ali
SAIF ALDIN ALI MADIN
سيف الدين علي ماضي
S96aif@gmail.com
Experiment Name:- Deflection of curved beam
2. Introduction
The deflection of a beam or bars must be often be limited in order to provide
integrity and stability of structure or machine. Plus, code restrictions often require
these members not vibrate or deflect severely in order to safely support their
intended loading.
This experiment helps us to show some kind of deflection and how to calculate the
deflection value by using Castigliano’s Theorem and make a comparison between
result of the experiment and the theory.
Deflection of curved beam |Strength of Material LaboratorySaif al-din ali
SAIF ALDIN ALI MADIN
سيف الدين علي ماضي
S96aif@gmail.com
Experiment Name:- Deflection of curved beam
2. Introduction
The deflection of a beam or bars must be often be limited in order to provide
integrity and stability of structure or machine. Plus, code restrictions often require
these members not vibrate or deflect severely in order to safely support their
intended loading.
This experiment helps us to show some kind of deflection and how to calculate the
deflection value by using Castigliano’s Theorem and make a comparison between
result of the experiment and the theory.
Project on Transformer Design | Electrical Machine DesignJikrul Sayeed
Transformer Design | Core Design | Full Design | EE 3220 Electrical Machine Design
EE-3220
Core Design
Window Dimensions
Yoke Design
Overall Dimensions of Frame
Low Voltage Winding
High Voltage Winding
Resistance
Leakage Reactance
Regulation
Losses
Core Loss
Efficiency
No Load Current
Tank
Project on Transformer Design
Roof Truss Design (By Hamza Waheed UET Lahore )Hamza Waheed
This presentation defines, describes and presents the most effective and easy way to design a roof truss with all the necessary steps and calculations based on Allowable Stress Design. Soft-wares like MD Solids, Truss Analysis have been used. It is most convenient way to design a roof truss which is being the most important structural components of All types of steel bridges.
Gantry girder
Gantry girder or crane girder hand operated or electrically operated overhead cranes in industrial building such as factories, workshops, steel works, etc. to lift heavy materials, equipment etc. and carry them from one location to other , within the building
The GANTRY GIRDER spans between brackets attached to columns, which may either be of steel or reinforced concrete. Thus the span of gantry girder is equal to centre to centre spacing of columns. The rails are mounted on gantry girders.
Loads acting on gantry girder
Gantry girder, having no lateral support in its length (laterally unsupported) has to withstand the following loads:
1. Vertical loads from crane :
Self weight of crane girder
Hook load
Weight of crab (trolley)
2. Impact load from crane :
As the load is lifted using the crane hook and moved from one place to another, and released at the required place, an impact is felt on the gantry girder.
3. Longitudinal horizontal force (Drag force) :
This is caused due to the starting and stopping of the crane girder moving over the crane rails, as the crane girder moves longitudinally, i.e. in the direction of gantry girder.
This force is also known as braking force, or drag force.
This force is taken equal to 5% of the static wheel loads for EOT or hand operated cranes.
4. Lateral load (Surge load) :
Lateral forces are caused due to sudden starting or stopping of the crab when moving over the crane girder.
Lateral forces are also caused when the crane is dragging weights across the' floor of the shop.
Types of gantry girders
Depending upon the span and crane capacity, there can be many forms of gantry girders. Some commonly used forms are shows in fig .
Rolled steel beams with or without plates, channels or angles are normally used for spans up to 8m and for cranes up to 50kN capacity.
Plate girder are suitable up to span 6 to 10 m.
Plate girder with channels, angles, etc. can be used for spans more than 10m
Box girder are used foe spans more than 12m.
Saudi Arabia stands as a titan in the global energy landscape, renowned for its abundant oil and gas resources. It's the largest exporter of petroleum and holds some of the world's most significant reserves. Let's delve into the top 10 oil and gas projects shaping Saudi Arabia's energy future in 2024.
TECHNICAL TRAINING MANUAL GENERAL FAMILIARIZATION COURSEDuvanRamosGarzon1
AIRCRAFT GENERAL
The Single Aisle is the most advanced family aircraft in service today, with fly-by-wire flight controls.
The A318, A319, A320 and A321 are twin-engine subsonic medium range aircraft.
The family offers a choice of engines
Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
Explore the innovative world of trenchless pipe repair with our comprehensive guide, "The Benefits and Techniques of Trenchless Pipe Repair." This document delves into the modern methods of repairing underground pipes without the need for extensive excavation, highlighting the numerous advantages and the latest techniques used in the industry.
Learn about the cost savings, reduced environmental impact, and minimal disruption associated with trenchless technology. Discover detailed explanations of popular techniques such as pipe bursting, cured-in-place pipe (CIPP) lining, and directional drilling. Understand how these methods can be applied to various types of infrastructure, from residential plumbing to large-scale municipal systems.
Ideal for homeowners, contractors, engineers, and anyone interested in modern plumbing solutions, this guide provides valuable insights into why trenchless pipe repair is becoming the preferred choice for pipe rehabilitation. Stay informed about the latest advancements and best practices in the field.
Forklift Classes Overview by Intella PartsIntella Parts
Discover the different forklift classes and their specific applications. Learn how to choose the right forklift for your needs to ensure safety, efficiency, and compliance in your operations.
For more technical information, visit our website https://intellaparts.com
Courier management system project report.pdfKamal Acharya
It is now-a-days very important for the people to send or receive articles like imported furniture, electronic items, gifts, business goods and the like. People depend vastly on different transport systems which mostly use the manual way of receiving and delivering the articles. There is no way to track the articles till they are received and there is no way to let the customer know what happened in transit, once he booked some articles. In such a situation, we need a system which completely computerizes the cargo activities including time to time tracking of the articles sent. This need is fulfilled by Courier Management System software which is online software for the cargo management people that enables them to receive the goods from a source and send them to a required destination and track their status from time to time.
Democratizing Fuzzing at Scale by Abhishek Aryaabh.arya
Presented at NUS: Fuzzing and Software Security Summer School 2024
This keynote talks about the democratization of fuzzing at scale, highlighting the collaboration between open source communities, academia, and industry to advance the field of fuzzing. It delves into the history of fuzzing, the development of scalable fuzzing platforms, and the empowerment of community-driven research. The talk will further discuss recent advancements leveraging AI/ML and offer insights into the future evolution of the fuzzing landscape.
About
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
1. Theory of Machines
By Khurmi
By Darawan Abdulwahid summer | 2017 darawan.me@gmail.com
Solution Manual | Chapter 11 |Belt, Rope and Pulley
2. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
1
N1 = 120 r. p. m d1 = 2m t = 5mm = 0.005m d2 = 1m N2 =?
Solution:
1.With no slip
N2
N1
=
d1 + t
d2 + t
→
N2
120
=
2 + 0.005
1 + 0.005
→ N2 = (
2 + 0.005
1 + 0.005
) × 120 = 239.4 r. p. m
2.With a slip of 3%
N2
N1
=
d1 + t
d2 + t
× (1 −
s1 + s2
100
) →
N2
120
=
2 + 0.00
1 + 0.00
× (1 −
3 + 0
100
)
→ N2 = (
2 + 0.005
1 + 0.005
) × (1 −
3 + 0
100
) × 120 = 232.22 r. p. m
v = 600 (m/ min)× (
1min
60s
) = 10m/s μ = 0.3 θ = 160° ×
π
180
= 2.8rad
T = 700N = T1 P =?
Solution:
T1
T2
= eμθ
= e0.3×2.8
= 2.31 → T2 =
T1
2.31
When T1 = 700N → T2 =
700
2.31
= 303N
P = (T1 − T2) × v = (700 − 303) × 10 = 3983W = 3.983KW
1. An engine shaft running at 120 r.p.m. is required to drive a machine shaft by means of a belt. The
pulley on the engine shaft is of 2 m diameter and that of the machine shaft is 1 m diameter. If the
belt thickness is 5 mm ; determine the speed of the machine shaft, when 1. there is no slip ; and
2. there is a slip of 3%. [Ans. 239.4 r.p.m. ; 232.3 r.p.m.]
3. A pulley is driven by a flat belt running at a speed of 600 m/min. The coefficient of friction
between the pulley and the belt is 0.3 and the angle of lap is 160°. If the maximum tension in the
belt is 700 N ; find the power transmitted by a belt. [Ans. 3.983 kW]
3. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
2
b =? P = 7.5kW = 7500W d = 300mm = 0.3m N = 1600r. p. m μ = 0.22
θ = 210° ×
π
180
= 3.665rad T′
= 8N/mm width
Solution:
v =
πdN
60
=
π × 0.3 × 1600
60
= 25.13m/s
T1
T2
= eμθ
= e0.22×3.665
= 2.2396 … … . a
P = (T1 − T2) × v → 7500 = (T1 − T2) × 25.13
→ T1 − T2 =
7500
25.13
= 298.44N → T1 = 298.44 + T2 … … . b
sub a in b ∶
298.44 + T2
T2
= 2.2396 → T2 = 240.761N
T1 = 298.44 + 240.761 = 539.2N
b =
T1
T′
=
539.2
8
= 67.4mm
4. Find the width of the belt, necessary to transmit 7.5 kW to a pulley 300 mm diameter, if the
pulley makes 1600 r.p.m and the coefficient of friction between the belt and the pulley is 0.22.
Assume the angle of contact as 210° and the maximum tension in the belt is not to exceed 8
N/mm width. [Ans. 67.4 mm]
4. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
3
b = 100mm 𝑥 = 2.4m d1 = 450mm = 0.45m μ = 0.3
d2 = 300mm = 0.3m T′
= 14N/m N1 = 120 r. p. m P =?
Solution:
v =
πd1N1
60
=
π × 0.45 × 120
60
= 2.827m/s
α = sin−1
(
d1 − d2
2𝑥
) = sin−1
(
0.45 − 0.3
2 × 2.4
) = 1.79° → 2α = 2 × 1.79° = 3.58°
θ = (180 − 2α) ×
π
180
= (180 − 3.58) ×
π
180
= 3.079 rad
T1
T2
= eμθ
= e0.3×3.079
= 2.518 … … . a
b =
T1
T′
→ 100 =
T1
14
→ T1 = 100 × 14 = 1400N … … . b
sub a in b ∶
T1
T2
= 2.518 → T2 =
T1
2.518
=
1400
2.518
= 555.996N
P = (T1 − T2) × v = (1400 − 555.996) × 2.827
P = 2386W ≈ 2.39kW
5. An open belt 100 mm wide connects two pulleys mounted on parallel shafts with their centers
2.4 m apart. The diameter of the larger pulley is 450 mm and that of the smaller pulley 300 mm.
The coefficient of friction between the belt and the pulley is 0.3 and the maximum stress in the
belt is limited to 14 N/mm width. If the larger pulley rotates at 120 r.p.m., find the maximum
power that can be transmitted. [Ans. 2.39 kW]
5. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
4
b = 125mm = 0.125m t = 6mm = 0.006m d = 750mm 0.75m μ = 0.3
N = 500 θ = 150° ×
π
180
= 2.617rad ρ = 1Mg/m3
= 1000kg/ m3
σ = 2.75Mpa = 2.75 × 106
Pa P =?
Solution:
v =
πdN
60
=
π × 0.75 × 500
60
= 19.634m/s > 10m/s , so we take Tc in the acount
T1
T2
= eμθ
= e0.3×2.617
= 2.1926 … … . a
Tc = mv2
= 𝜌𝑏𝑡𝑙 × v2
= 1000 × 0.125 × 0.006 × 1 × (19.634)2
= 289.12N
T = σbt = 2.75 × 106
× 0.125 × 0.006 = 2062.5N/m2
T = Tc + T1 → T1 = T − Tc = 2062.5 − 289.12 = 1773.38 … … . b
sub a in b ∶
T1
T2
= 2.1926 → T2 =
T1
2.1926
=
1773.38
2.1926
= 808.8N
P = (T1 − T2) × v = (1773.38 − 808.8) × 19.634
P = 18931.1W ≈ 19kW
6. A leather belt 125 mm wide and 6 mm thick, transmits power from a pulley 750 mm diameter
which runs at 500 r.p.m. The angle of lap is 150° and μ = 0.3. If the mass of 1 m3 of leather is 1 Mg
and the stress in the belt is not to exceed 2.75 MPa, find the maximum power that can be
transmitted. [Ans. 19 kW]
6. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
5
P = 35kW = 35000W d = 1.5m N = 300 r. p. m θ =
11
24
× 2π = 2.88 rad
μ = 0.3 t = 9.5mm = 0.095m ρ = 1.1Mg/m3
= 1100 kg/m3
σ = 2.5Mpa = 2.5 × 106
Pa
Solution:
v =
πdN
60
=
π × 1.5 × 300
60
= 23.56m/s
T1
T2
= eμθ
= e0.3×2.88
= 2.37 … … . a
P = (T1 − T2) × v → 35000 = (T1 − T2) × 23.56
→ T1 − T2 =
35000
23.56
= 1485.568N → T1 = 1485.568 + T2 … … . b
sub a in b ∶
1485.568 + T2
T2
= 2.37 → T2 = 1081.39N
→ T1 = 1485.568 + 1081.39 = 2567.96N
Tc = mv2
= 𝜌𝑏𝑡𝑙 × v2
= 1100 × b × 0.095 × 1 × (23.56)2
= 5800.5b
Tc + T1 = σbt → 5800.5b + 2566.96 = 2.5 × 106
× b × 0.095
2566.96 = 237500b − 5800.5b → 2566.96 = 17950b
b =
2566.96
17950
= 0.143 m = 143 mm
P = (T1 − T2) × v = (2566.96 − 17950) × 23.56
P = 35920W ≈ 35.9 kW
7. A flat belt is required to transmit 35 kW from a pulley of 1.5 m effective diameter running at
300 r.p.m. The angle of contact is spread over 11/24 of the circumference and the coefficient of
friction between belt and pulley surface is 0.3. Determine, taking centrifugal tension into
account, width of the belt required. It is given that the belt thickness is 9.5 mm, density of its
material is 1.1 Mg/m3 and the related permissible working stress is 2.5 MPa. [Ans. 143 mm]
7. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
6
N1 = 750r. p. m t = 8mm = 0.008m b = 250mm = 0.25m μ = 0.35
d1 = 350mm = 0.35m d2 = 1350mm = 1.35m 𝑥 = 1350mm = 1.35m
σ = 2.5N/mm2
× 106
= 2.5 × 106
Pa m = 2kg/m P = ?
Solution:
v =
πd1N1
60
=
π × 0.35 × 750
60
= 13.74m/s
Tc = mv2
= 2 × (13.74)2
= 377.82N
sinα = (
d1 − d2
2𝑥
) → α = sin−1
(
0.34 − 1.35
2 × 1.35
) = | − 21.73°| = 21.73° Must be + ive
θ = (180 − 2α) ×
π
180
= (180 − (2 × 21.73)) ×
π
180
= 2.38 rad
T1
T2
= eμθ
= e0.35×2.38
= 2.302 … … . a
T = σbt = Tc + T1 → 2.5 × 106
× 0.25 × 0.008 = 377.82 + T1
5000 − 377.82 = T1 → T1 = 4622.18N … … . b
sub a in b ∶
4622.18
T2
= 2.302 → T2 = 2007.89N
P = (T1 − T2) × v = (4622.18 − 2007.89) × 13.74
P = 35920.34W ≈ 35.9kW
8. A blower is driven by an electric motor though a belt drive. The motor runs at 750 r.p.m. For this
power transmission, a flat belt of 8 mm thickness and 250 mm width is used. The diameter of the
motor pulley is 350 mm and that of the blower pulley 1350 mm. The center distance between these
pulleys is 1350 mm and an open belt configuration is adopted. The pulleys are made out of cast
iron. The frictional coefficient between the belt and pulley is 0.35 and the permissible stress for the
belt material can be taken as 2.5 N/mm² with sufficient factor of safety. The mass of a belt is 2 kg
per meter length. Find the maximum power transmitted without belt slipping in any one of the
pulleys. [Ans. 35.9 kW]
8. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
7
d1 = 1.2m d2 = 0.5m 𝑥 = 3.6m m = 1kg/m T = 2kN = 2000N
N1 = 200r. p. m N2 = 450r. p. m μ = 0.3 1. Tq1 & Tq2 =?
2. P =? 3. Power lost in friction =? Efficiency =?
Solution:
v =
πd1N1
60
=
π × 1.2 × 200
60
= 12.566m/s
α = sin−1
(
d1 − d2
2𝑥
) = sin−1
(
1.2 − 0.5
2 × 3.6
) = 5.579°
θ = (180 − 2α) ×
π
180
= (180 − (2 × 5.579)) ×
π
180
= 2.946 rad
T1
T2
= eμθ
= e0.3×2.946
= 2.42 … … . a
Tc = mv2
= 1 × (12.566)2
= 157.91N
T = Tc + T1 → 2000 = 157.91 + T1 → T1 = 2000 − 157.91 = 1842.09N … … . b
sub a in b ∶
1842.09
T2
= 2.42 → T2 =
1842.09
2.42
= 761.194N
2.
P = (T1 − T2) × v = (1842.09 − 761.194) × 12.556 = 13588W = 13.588kW
9. An open belt drive connects two pulleys 1.2 m and 0.5 m diameter on parallel shafts 3.6 m apart.
The belt has a mass of 1 kg/m length and the maximum tension in it is not to exceed 2 kN. The 1.2
m pulley, which is the driver, runs at 200 r.p.m. Due to the belt slip on one of the pulleys, the
velocity of the driven shaft is only 450 r.p.m. If the coefficient of friction between the belt and the
pulley is 0.3, find : 1. Torque on each of the two shafts, 2. Power transmitted, 3. Power lost in
friction, and 4. Efficiency of the drive.
[Ans. 648.6 N-m, 270.25 N-m ; 13.588 kW ; 0.849 kW ; 93.75%]
9. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
8
1. Tq = (T1 − T2) × r
Tq1 = (T1 − T2) × r1 = (1842.09 − 761.194) ×
1.2
2
= 648.53N. m
Tq2 = (T1 − T2) × r2 = (1842.09 − 761.194) ×
0.5
2
= 270.22N. m
3. Power lost in friction = input power − output power
input power =
2πN1Tq1
60
=
2π × 200 × 648.53
60
= 13.582kW
output power =
2πN2Tq2
60
=
2π × 450 × 270.22
60
= 12.733kW
Power lost in friction = 13.582 − 12.733 = 0.848kW
4.
Efficiency =
output power
input power
× 100% =
12.733
13.582
× 100% = 93.75%
10. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
9
𝑥 = 3.5m d1 = 600mm = 0.6m d2 = 300mm = 0.3m P = 6kW = 6000W
N1 = 220r. p. m T′
= 25N/m width t = 5mm = 0.005m μ = 0.35
1. L =? 2. b =? 3. Tₒ =?
Solution:
1. L =
π
2
(d1 + d2) + 2𝑥 +
(d1−d2)2
4𝑥
=
π
2
(0.6 + 0.3) + (2 × 3.5) +
(0.6+0.3)2
4×3.5
= 8.472m
2.
v =
πd1N1
60
=
π × 0.6 × 220
60
= 6.911m/s
α = sin−1
(
0.6 + 0.3
2 × 3.5
) = sin−1
(
0.6 + 0.3
2 × 3.5
) = 7.387°
θ = (180 + 2α) ×
π
180
= (180 + (2 × 7.387)) ×
π
180
= 3.399 rad
T1
T2
= eμθ
= e0.3×3.399
= 3.286 … … . a
P = (T1 − T2) × v → 6000 = (T1 − T2) × 6.911
→ T1 − T2 =
6000
6.911
= 868.181N → T1 = 868.181 + T2 … … . b
sub a in b ∶
868.181 + T2
T2
= 3.286 → T2 = 379.781N
T1 = 868.181 + 379.781 = 1247.926N
b =
T1
T′
=
1247.926
25
= 49.9mm
3.
Tₒ =
T1 + T2
2
=
1247.926 + 379.78
2
= 889.08
10. The power transmitted between two shafts 3.5 metres apart by a cross belt drive round the two
pulleys 600 mm and 300 mm in diameters, is 6 kW. The speed of the larger pulley (driver) is 220
r.p.m. The permissible load on the belt is 25 N/mm width of the belt which is 5 mm thick. The
coefficient of friction between the smaller pulley surface and the belt is 0.35. Determine :
1. necessary length of the belt ; 2. width of the belt, and 3. necessary initial tension in the belt.
[Ans. 8.472 m ; 53 mm ; 888 N]
11. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
10
T = 8mm = 0.008m b = 100mm = 0.1m v = 1600m/min ×
1min
60sec
= 26.67m/s
m = 0.9kg/m θ = 165° ×
π
180
= 2.88rad μ = 0.3 σ = 2MN/m2
= 2 × 106
pa
1. P =? 2. Tₒ =?
Solution:
1.
T1
T2
= eμθ
= e0.3×2.88
= 2.372 … … . a
Tc = mv2
= 0.9 × (26.67)2
= 640N
T = σ × b × t = Tc + T1 → 2 × 106
× 0.1 × 0.008 = 640 + T1
T1 = 1600 − 640 → T1 = 960N … … . b
sub a in b ∶
960
T2
= 2.372 → T2 = 404.72N
P = (T1 − T2) × v → P = (960 − 404.72) × 26.67 = 14809.3W = 14.8kW
2.
Tₒ =
T1 + T2 + 2Tc
2
=
960 + 404.72 + (2 × 640)
2
= 1322.36N
But when we use equation below , we get the same result which given for [Tₒ]∗
Tₒ =
T1 + T2 + Tc
2
=
960 + 404.72 + 640
2
= 1002.36N
-------------------------------------------------------------------------------------------------------------------------------
∗ see a textbook (theory of machines by r. k. bansal)
11. A flat belt, 8 mm thick and 100 mm wide transmits power between two pulleys, running at
1600 m/min. The mass of the belt is 0.9 kg/m length. The angle of lap in the smaller pulley is 165°
and the coefficient of friction between the belt and pulley is 0.3. If the maximum permissible stress
in the belt is 2 MN/m2, find : 1. maximum power transmitted ; and 2. initial tension in the belt
[Ans. 14.83 kW ; 1002 N]
12. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
11
d = 400mm = 0.4m N = 200r. p. m θ = 160° ×
π
180
= 2.79rad μ = 0.25
P = 3kW = 3000W
Solution:
v =
πdN
60
=
π × 0.4 × 200
60
= 4.188m/s
T1
T2
= eμθ
= e0.25×2.79
= 2 … … . a
P = (T1 − T2) × v → 3000 = (T1 − T2) × 4.188
→ T1 − T2 =
3000
4.188
= 716.33N → T1 = 716.33 + T2 … … . b
sub a in b ∶
716.33 + T2
T2
= 2 → T2 = 716.33N
T1 = 716.33 + 716.33 = 1432.66N
Tₒ =
T1 + T2
2
=
1432.66 + 716.33
2
= 1079.495N … … . e
12. An open belt connects two flat pulleys. The smaller pulley is 400 mm diameter and runs at 200
r.p.m. The angle of lap on this pulley is 160° and the coefficient of friction between the belt and
pulley face is 0.25. The belt is on the point of slipping when 3 kW is being transmitted. Which of the
following two alternatives would be more effective in order to increase the power :
1. Increasing the initial tension in the belt by 10 per cent, and
2. Increasing the coefficient of friction by 10 per cent by the application of a suitable dressing to the
belt? [Ans. First method is more effective]
13. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
12
1st Method , when Increasing the initial tension in the belt by 10%
Tₒ = 1074.495 +
1074.495
10
= 1181.94N
Tₒ =
T1 + T2
2
→ 1181.94 =
T1 + T2
2
→ T1 + T2 = 2363.896N
T1 = 2363.896 − T2 … … . c
sub c in a ∶
2363.896 − T2
T2
= 2 → T2 = 787.965N
T1 = 2363.896 − 787.965 = 1575.93N
P = (T1 − T2) × v → P = (1575.93 − 787.965) × 4.188 = 3300W = 3.3kW
2nd Method , when Increasing the coefficient of friction in the belt by 10%
μ = 0.25 +
0.25
10
= 0.275
T1
T2
= eμθ
= e0.275×2.79
= 2.15 → T1 = 2.15 × T2 … … . d
Tₒ =
T1 + T2
2
→
T1 + T2
2
= 1079.495
T1 + T2 = 2 × 1079.495 = 2158.99 … … . f
T2 = 620.825 & T1 = 1334.744
P = (T1 − T2) × v → P = (1334.744 − 620.825) × 4.188 = 2989W = 2.99kW
P at 1st method is grater than the 2nd So , First method is more effective
14. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
13
m = 0.9kg/m T = 2.2kN = 2200N θ = 170° ×
π
180
= 2.967rad μ = 0.17
2β = 45° 1. v =? 2. P =?
Solution:
For maximum power we can write :
1.
v = √
T
3m
= √
2200
3 × 0.9
= 28.54m/s
T1
T2
= e
μθ×
1
sinβ = 𝑒0.17×2.967×
1
sin22.5 = 3.73 … … . a
Tc = mv2
= 0.9 × (28.54)2
= 733N
T1 = T − Tc = 2200 − 733 = 1466.67N … … . b
sub a in b ∶
1466.67
T2
= 3.73 → T2 =
1466.67
3.73
= 393.2N
2.
P = (T1 − T2) × v → P = (1466.67 − 393.2) × 28.54 = 30636.833W = 3.7kW
14. Power is transmitted between two shafts by a V-belt whose mass is 0.9 kg/m
length. The maximum permissible tension in the belt is limited to 2.2 kN. The angle of
lap is 170° and the groove angle 45°. If the coefficient of friction between the belt
and pulleys is 0.17, find : 1. velocity of the belt for maximum power ; and 2. power
transmitted at this velocity. [Ans. 28.54 m/s ; 30.7 kW]
15. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
14
𝑥 = 1m P = 100kW d1 = 300mm = 0.3m N1 = 1000 r. p. m
N2 = 375 r. p. m 2β = 40° A = 400mm2
= 4 × 10−4
m2
σ = 2.1Mpa = 2.1 × 106
Pa ρ = 1100kg/m3
μ = 0.28 n =?
Solution:
N2
N1
=
d1
d2
→ d2 =
N1 × d1
N2
=
1000 × 0.3
375
= 0.8m
α = sin−1
(
d1 − d2
2𝑥
) = sin−1
(
0.3 − 0.8
2 × 1
) = | − 14.477°| = 14.477°
θ = (180 − 2α) ×
π
180
= (180 − (2 × 14.477°)) ×
π
180
= 2.636 rad
T1
T2
= e
μθ×
1
sinβ = 𝑒0.28×2.636×
1
sin20 = 8.655 … … . a
T = σ × A = 2.1 × 106
× 4 × 10−4
= 840N
v =
πd1N1
60
=
π × 0.3 × 1000
60
= 15.7m/s
Tc = ρAv2
= 1100 × 4 × 10−4
× (15.7)2
= 108.565N
T1 = T − Tc = 840 − 108.565 = 731.43N … … . b
sub a in b ∶
T1
T2
= 8.655 → T2 =
731.43
8.655
= 85.393N
P = (T1 − T2) × v → P = (731.43 − 85.393) × 15.7 = 10142.7 = 10.14kW
n =
Total power
Power per belt
=
100
10.14
= 9.86 = 10 belts
15. Two shafts whose centers are 1 m apart are connected by a V-belt drive. The driving pulley is
supplied with 100 kW and has an effective diameter of 300 mm. It runs at 1000 r.p.m. while the
driven pulley runs at 375 r.p.m. The angle of groove on the pulleys is 40°. The permissible tension in
400 mm² cross-sectional area belt is 2.1 MPa. The density of the belt is 1100 kg/m³. The coefficient
of friction between the belt and pulley is 0.28. Estimate the number of belts required. [Ans. 10]
16. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
15
P = 230kW = 230 × 103
W d = 1m N = 450 r. p. m T = 800N
m = 0.46
kg
meter
θ = 160° ×
π
180
= 2.79rad 2β = 45° μ = 0.3 n =?
Solution:
v =
πdN
60
=
π × 1 × 450
60
= 23.56m/s
T1
T2
= e
μθ×
1
sinβ = 𝑒0.3×2.79×
1
sin22.5 = 8.9 … … . a
Tc = mv2
= 0.46 × (23.56)2
= 255.33N
T1 = T − Tc = 800 − 255.33 = 544.67N … … . b
sub a in b ∶
T1
T2
= 8.9 → T2 =
544.67
8.9
= 61.2N
P = (T1 − T2) × v → P = (544.67 − 61.2) × 23.56 = 11390.5W = 11.39kW
n =
Total power
Power per belt
=
230
11.39
= 20.2 = 21 belts
16. A rope drive is required to transmit 230 kW from a pulley of 1 metre diameter running at 450
r.p.m. The safe pull in each rope is 800 N and the mass of the rope is 0.46 kg per metre length. The
angle of lap and the groove angle is 160° and 45° respectively. If the coefficient of friction between
the rope and the pulley is 0.3, find the number of ropes required. [Ans. 21]
17. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
16
𝑥 = 3m d1 = 3m d2 = 2m 2β = 40° m = 3.7kg per meter μ = 0.15
T = 20kN = 20000N P =? N2 =?
Solution:
α = sin−1
(
d1 − d2
2𝑥
) = sin−1
(
3 − 2
2 × 3
) = 14.477°
θ = (180 − 2α) ×
π
180
= (180 − (2 × 14.477°)) ×
π
180
= 2.636 rad
T1
T2
= e
μθ×
1
sinβ = 𝑒0.15×2.636×
1
sin20 = 3.177 … … . a
For maximum power we can write :
Tc =
T
3
=
20000
3
= 6666.67N
v = √
T
3m
= √
20000
3 × 3.7
= 42.447m/s
T1 = T − Tc = 20000 − 6666.67 = 13333.33N … … . b
sub a in b ∶
T1
T2
= 3.177 → T2 =
13333.33
3.177
= 4196.8N
P = (T1 − T2) × v → P = (13333.33 − 4196.8) × 42.447 = 387818.28W = 387.822kW
v =
πd2N2
60
→ N2 =
v × 60
π × d2
=
42.447 × 60
π × 2
= 405.3 r. p. m
17. Power is transmitted between two shafts, 3 metres apart by an open wire rope passing round
two pulleys of 3 metres and 2 metres diameters respectively, the groove angle being 40°. If the
rope has a mass of 3.7 kg per metre length and the maximum working tension in rope is 20 kN,
determine the maximum power that the rope can transmit and the corresponding speed of the
smaller pulley. The coefficient of friction being 0.15. [Ans. 400 kW ; 403.5 r.p.m.]
18. Solution manual - chapter 11 Preapred by Darawan Abdulwahid
17
P = 75kW = 75000W d1 = 1.5m 2β = 45° N1 = 200 r. p. m μ = 0.3
θ = 160° ×
π
180
= 2.79rad m = 0.6kg per meter T = 800N n =? Tₒ =?
Solution:
v =
πd1N1
60
=
π × 1.5 × 200
60
= 15.7m/s
T1
T2
= e
μθ×
1
sinβ = 𝑒0.3×2.79×
1
sin22.5 = 8.91 … … . a
Tc = mv2
= 0.6 × (15.7)2
= 148.04N
T1 = T − Tc = 800 − 148.04 = 651.955N … … . b
sub a in b ∶
T1
T2
= 8.91 → T2 =
651.955
8.91
= 73.056N
P = (T1 − T2) × v → P = (651.955 − 73.056) × 15.7 = 9088.714W = 9.09kW
n =
Total power
Power per belt
=
75
9.09
= 8.25 = 9 belts
-------------------------------------------------------------------------------------------------------------------------------
Summer 2016 _Edited 2017
Contact me
Gmail : Darawan .me @gmail.com
18. A rope drive transmits 75 kW through a 1.5 m diameter, 45° grooved pulley rotating at 200
r.p.m. The coefficient of friction between the ropes and the pulley grooves is 0.3 and the angle of
lap is 160°. Each rope has a mass of 0.6 kg/m and can safely take a pull of 800 N. Taking centrifugal
tension into account determine : 1. the number of ropes required for the drive, and 2. initial rope
tension. [Ans. 9 ; 510.2 N]