Uploaded bymechestud
8,076 views

J3010 Unit 6

The document discusses different types of belting used to transmit power between rotating shafts in factories, including flat belts and V-belts. It provides objectives and formulas for calculating the length of open and closed belt drives, as well as the power transmitted by a belt based on the tension in the tight and slack sides and the belt velocity. Worked examples are included to demonstrate calculating belt length and transmitted power.

Downloaded 293 times
BELTING J 3010/6/1 UNIT 6 BELTING OBJECTIVES General Objective : To understand and apply the concept of belting Specific Objectives : At the end of this unit you will be able to:  state the difference between open and close belt.  explain that power transmitted by the flat belt and V belt.  explain that ratio tension for flat and V belt.  calculate power transmitted by the belt with consider centrifugal force.
BELTING J 3010/6/2 INPUT 6.0 INTRODUCTION In factories, the power has to be transmitted from one shaft to another, then belt driving between pulleys on the shafts may be used. The pulley rotating shaft is called driver. The pulley intended to rotate is known, as follower or driven. When the driver rotates, it carries the belt that grip between its surface and the belt. The belt, in turn, carries the driven pulley which starts rotating. The grip between the pulley and the belt is obtained by friction, which arises from pressure between the belt and the pulley. (a) Types of belts The flat belt is mostly used in the factories and work shops, where a moderate amount of power is to be transmitted, from one pulley to another, when the two pulleys are not more than 10 m a part. The V-belt is mostly used in the factories and work shops where a great amount of power is to be transmitted, from one pulley to another, when the two pulleys are very near to each other.
BELTING J 3010/6/3 The types of belts:- a. flat belt b. V belt 6.1 LENGTH OF AN OPEN BELT DRIVE A B K α1 α2 C O1 O2 F r2 r1 D E d Fig.6.1 Open belt drive O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = Distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle BO2C = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian)
BELTING J 3010/6/4 We know that the length of the belt, L = Arc AFE + ED + Arc DCB + BA r1  r2 Cos α1 = d r r α1 = cos-1 1 2 d θ1 = 2π - 2α1 = 2 (π - α1) (radian) θ2 = α1 = 2 α2 (radian) Arc AFE = r1θ1 Arc DCB = r2θ2 And ED = BA = KO2 KO2 Sin α1 = d KO2 = d Sin α1 Finally the total of length of belt, L = Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sin α1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1
BELTING J 3010/6/5 Example 6.1 Find the length of belt necessary to drive a pulley of 480 cm diameter running parallel at a distance of 12 meter from the driving pulley of diameter 80 cm. This system is an open belt drive. Solution 6.1 A B K F α1 O2 α2 C O1 r2 r1 D E d Fig.6.2 Open belt drive O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = Distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle BO2C = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian) Radius of smaller pulley = 80/2 = 40 cm. Radius of larger pulley = 480/2 = 240 cm. Distance between the pulleys, d = 12m = 1 200 cm We know that the length of belt is, L = Arc AFE + ED + Arc DCB + BA r1  r2 240  40 Cos α1 = = d 1200
BELTING J 3010/6/6 240  40 α1 = cos-1 1200 = cos-1 0.16667 = 80º α1 = 1.396 radian θ1 = 2π - 2α1 = 2 (π - 1.396 ) (radian) = 3.491 radian θ2 = 2α1 = 2 α2 (radian) = 2 (1.396 ) = 2.792 radian Arc AFE = r1θ1 = 240 x 3.491 = 837.84 cm Arc DCB = r2θ2 = 40 x 2.792 = 111.68 cm And ED = BA = KO2 KO2 Sin 80º = d KO2 = d Sin α1 = 1 200 x 0.98481 = 1181.77 m Finally the total of belt length is L = Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sin α1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1 = 240 x 3.491 + 40 x 2.792 +2 x 1181.77 = 837.84 + 111.68 + 2363.54 = 3313.06 cm = 33.13 m
BELTING J 3010/6/7 6.2 LENGTH OF CLOSE BELT DRIVE K A D C O1 α1 α2 O2 F r2 r1 B E d Fig.6.3 Cross belt drive O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = Distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle DO2O1 = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian) We know that the length of the belt is L = Arc AFE + ED + Arc DCB + BA r1  r2 Cos α1 = d r r α1 = cos-1 1 2 d θ1 = 2π - 2α1 = 2 (π - α1) (radian) θ1 = θ2
BELTING J 3010/6/8 Arc AFE = r1θ1 Arc DCB = r2θ2 And ED = BA = KO2 KO2 Sin α1 = d KO2 = d Sin α1 Finally the total of length belt, L = Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sin α1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1 Example 6.2 Find the length of belt for a cross belt drive system. The diameter of the drive pulley is 480 cm which running parallel at a distance of 12 meter from the driving pulley which has a diameter of 80 cm. Solution 6.2 K A D α1 α2 C F O1 O2 r2 r1 B E d Fig.6.4 Cross belt drive
BELTING J 3010/6/9 O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle DO2O1 = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian) Radius of smaller pulley = 80/2 = 40 cm. Radius of larger pulley = 480/2 = 240 cm. Distance between the pulleys, d = 12m = 1 200 cm We know that the length of the belt, L = Arc AFE + ED + Arc DCB + BA r1  r2 240  40 Cos α1 = = d 1200 240  40 α1 = cos-1 1200 -1 = cos 0.23333 = 76.5º α1 = 1.335 radian θ1 = 2π - 2α1 = 2 (π - α1) (radian) = 2 (π - 1.335) (radian) = 3.613 radian θ1 = θ2 Arc AFE = r1θ1 = 240 x 3.613 = 867.12 cm Arc DCB = r2θ2 = 40 x 3.613 = 144.52 cm And ED = BA = KO2 KO2 Sin α1 = d KO2 = d Sin 76.5º =1 200 x 0.9724 ED = BA = KO2 = 1166.88 cm
BELTING J 3010/6/10 Finally the total of length belt, L= Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sinα1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1 = 240 x 3.613 + 240 x 3.613 + 2 x 1 200 Sin 76.5º = 867.12 + 144.52 + 2333.76 = 3345.4 cm L = 33.45 m
BELTING J 3010/6/11 Activity 6A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 6.1 Two pulleys, one with a 450 mm diameter and the other with a 200 mm diameter are on parallel shaft of 1.95 m apart and are connected by a cross belt. Find the length of the belt required and the angle of contact between the belt and each pulley. 6.2 It is required to drive a shaft B at 620 rpm by means of a belt from a parallel shaft A. A pulley of 30 cm diameter on shaft A runs at 240 rpm. Determine the size of pulley on the shaft B. 6.3 Find the length of belt necessary to drive a pulley of 1.4 m diameter running parallel at a distance of 1.7 meter from the driving pulley of diameter 0.5 m. It is connected by an open belt. N1 d 2  N 2 d1 N1 = speed diver in rpm N2 = speed follower in rpm d1 ,d2 = diameter pulley driver and follower.
BELTING J 3010/6/12 Feedback to Activity 6A Have you tried the questions????? If “YES”, check your answers now 6.1 4.975 m; 199 or 3.473 radian. 6.2. 11.6 cm 6.3 6.51 m
BELTING J 3010/6/13 INPUT 6.3 POWER TRANSMITTED BY A BELT Power = (T1 –T2) v Where, T1 =Tension in tight side in Newton. T2 = Tension in slack side. V = velocity of belt Fig. 6.5, shows the driving pulley (i.e., driver) A and the follower B. The driving pulley pulls the belt from one side, and delivers the same to the other.The maximum tension in the tight side will be greater than that slack side. Fig. 6.5 Torque exerted on the driving pulley = ( T1 - T2) r1 Torque exerted on the driven or follower = ( T1 - T2) r2 Power transmitted = Force x distance = ( T1 - T2) v
BELTING J 3010/6/14 Example 6.3 The tension in the two sides of the belt are 100 N and 80 N respectively. If the speed of the belt is 75 meters per second. Find the power transmitted by the belt. Solution 6.3 Given, Tension in tight side, T1 = 100 N Tension in slack side, T2 = 80 N Velocity of belt, v = 75 m/s Let P = Power transmitted by the belt Using the relation, Power = (T1 –T2) v = ( 100 – 80) 75 = 1500 watt = 1.5 kw.
BELTING J 3010/6/15 Activity 6B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 6.4. Find the tension in the tight side, if the tension in slack side 150 N. The speed of the belt is 58 meters per second and the power transmitted by this belt is 2 kw. 6.5 Diameter of driver is 50 mm. If the driver transmits 8 kw when it is rotating at 300 rev/ min. Calculate velocity of driver. 6.6 The tension in the two sides of the belt are 300 N and 180 N respectively. If the speed of the belt is 50 meters per second, find the initial tension and power transmitted by the belt. 6.7 A 100 mm diameter pulley is belt-driven from a 400 mm diameter pulley. The 400 mm pulley rotates at 480 rev/min. The number of rev/sec of the 100 mm diameter pulley is A. 2 B. 32 C. 120 D. 1920 T1  T2 T0 = 2 Where, T0 = Initial tension in the belt.
BELTING J 3010/6/16 Feedback to Activity 6B Have you tried the questions????? If “YES”, check your answers now 6.4 115.5 N 6.5 0.785 m/s 6.6 6 000 watt or 6 kw; 240 N 6.7 B. 32
BELTING J 3010/6/17 INPUT 6.4 RATIO OF TENSIONS. Fig. 6.6 Consider a driven pulley rotating in the clockwise direction as in fig 6.6. Let T1 = Tension in the belt on the tight side. T2 = Tension in the belt on the slack side. ө = An angle of contact in radians (i.e , angle subtended by the arc AB, along which the belt touches the pulley, at the centre) Now consider a small position of the belt PQ, subtending an angle  at the centre of the pulley as shown fig 6.6. The belt PQ is in equilibrium under the following force: i. Tension T in the belt at P. ii. Tension T + T in the belt at Q. iii. Normal reaction R, and iv. Frictional force F =  x R Where  is the coefficient of friction between the belt and pulley. Resolving all the forces horizontally and equating the same,
BELTING J 3010/6/18   R = ( T + T) sin + T sin (6.1) 2 2 Since the  is very small, therefore, substituting,   sin = in equation 6.1 2 2   R = ( T + T) +T 2 2 T  T  T . = + + 2 2 2 T  = T. ( neglecting ) (6.2) 2 Now resolving the forces vertically,    x R = (T + T) cos - T cos (6.3) 2 2  Since the angle . is very small, therefore substituting cos = 1 in 2 equation 6.3 or, xR = T + T – T = T T R = (6.4)  Equating the values of R from equation 6.2 and 6.4, T Or T. =  T Or = . T Integrating both sides from A to B,  T T1 T T2     0 T1 or loge =  T2 T1 = e  ration of tension for flat belt T2 T1 = e /sin  ration of tension for V belt T2 where  = half angle of groove
BELTING J 3010/6/19 Example 6.4 Find the power transmitted by a belt running over a pulley of 60 cm diameter at 200 rpm. The coefficient of friction between the pulley is 0.25, angle of lap 160 and maximum tension in the belt is 250 KN. Solution 6.4 Given, Diameter of pulley, d = 60 cm = 0.6 m Speed of pulley N = 200 rpm  dN  x 0.6 x 200 Speed of belt, v = = = 2 = 6.28 m/ sec 60 60 Coefficient of friction, μ = 0.25 160 x  Angle of contact,  = 160 = = 2.7926 radian. 180 Maximum tension in the belt, T1 = 250 kN Let P = power transmitted by the belt. Using the relation, T1 = e  T2 = e 0.25 x 2.7926 T1 = 2.01 T2 T1 250 T2 = = = 124.38 kN 2.01 2.01 Now using the relation, P = (T1 –T2 ) v = ( 250 - 124.38 ) 6.28 = 788.89 watt P = 0.79 kw
BELTING J 3010/6/20 INPUT 6.5 CENTRIFUGAL TENSION. Fig.6.7 The tension caused by centrifugal force is called centrifugal tension. At lower speeds the centrifugal tension is very small, but at higher speeds its effect is considerable, and thus should be taken into account. Consider a small portion PQ of the belt subtending an angle d at the centre of the pulley as show in fig 6.7. Let M = mass of the belt per unit length, V = linear velocity of the belt, r = radius of the pulley over which the belt runs, Tc = Centrifugal tension acting tangentially at P and Q. Length of the belt PQ, = r d Mass of the belt PQ, M = M r d We know that centrifugal force, = M v2/r
BELTING J 3010/6/21 Centrifugal force of the belt PQ M x r.d v 2 = r = M x d v 2 The centrifugal tension (Tc) acting tangentially at P and Q keeps the belt in equilibrium. Now resolving the force (i.e, centrifugal force and centrifugal tension) horizontally and equating the same, d 2 Tc sin = M x d v 2 2 since angle d is very small, therefore, substituting d d sin = 2 2 d 2 Tc = M x d v 2 2 Tc = M x d v 2 / d Tc = M v2 T1  Tc  e ratio tension for flat belt. T2  Tc T1  Tc  e   / sin  ratio tension for V belt T2  Tc 6.5.1 Condition for the transmission of maximum power The maximum power, When T c = ⅓T1 It shows that the power transmitted is maximum ⅓ of the maximum tension is absorbed as centrifugal tension. The velocity of belt for maximum transmission of power may be obtained from equation T1 = 3Tc = M v2.
BELTING J 3010/6/22 3 T1 v2 = M T1 v = 3M Example 6.5 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 0.9 kg/m length and the maximum tension in it is not exceed 2 kN. The larger pulley runs at 200 rev/min. Calculate the torque on each of the two shafts and the power transmitted. Coefficient of friction is 0.3 and angle of lap on the smaller pulley is 168° ( 2.947 radian ). Solution 6.5 Given Diameter of larger pulley, d1 = 1.2 m Radius of the larger pulley, r1 = 0.6 Diameter of smaller pulley, d2 =0.5 m Radius of the smaller pulley, r2 = 0.25 m Distance between two shaft, D = 3.6 m Mass of the belt per meter length, M = 0.9 kg/m Maximum Tension, T1 = 2 kN = 2 000 N Speed of the larger pulley = 200 rpm Velocity of the belt,  x 1.2 x 200 V= = 12.57 m/s 60 T c = Mv2 = 0.9 x 12.572 = 142.2 N
BELTING J 3010/6/23 e µθ = e 0.3 x 2.947 = 2.421 Using the relation, T1  Tc  e T2  Tc 2000  142.2 = 2.421 T2  142.2 (T2 – 142.2) 2.421 = 1857.8 1857.8 (T2 – 142.2) = = 767.37 2.421 T2 = 767.37 + 142.2 T2 = 909.57 N We know that the torque on the larger pulley shaft (TL), TL = ( T1 – T2) r1 = ( 2000 – 909.57) 0.6 = 654.26 Nm. Torque on the smaller pulley shaft (Ts), Ts = ( T1 – T2) r2 = (2000 – 909.57) 0.25 = 272.61 Nm. Power transmitted, P = ( T1 – T2)v = ( 2000 – 909.57) 12.57 = 13706.71 watt = 13.71 kw
BELTING J 3010/6/24 Activity 6C TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT INPUT…! 6.8 Two pulleys, one with a 450 mm diameter and the other with a 200 mm diameter are on parallel of shafts 1.95 m apart. What power can be transmitted by the belt when the larger pulley rotates at 200 rev/min, if the maximum permissible tension in the belt is 1 kN and the coefficient of friction between the belt and pulley is 0.25. Angle of contact between the belt and larger pulley is 3.477 radian. 6.9 Find the power transmitted by a V drive from the following data: Angle of contact = 84º Pulley groove angle = 45º Coefficient of friction = 0.25 Mass of belt per meter length = 0.472 kg/m Permissible tension = 139 N Velocity of V belt = 12.57 m/s. 6.10 An open belt drive connects two pulleys 1.2 m and 0.6 m, on parallel shafts 3 m apart. The belt has a mass 0f 0.56 kg/m and maximum tension is 1.5 kN. The driver pulley runs at 300 rev/min. Calculate the inertial tension, power transmitted and maximum power. The coefficient of friction between the belt and the pulley surface is 0.3.
BELTING J 3010/6/25 Feedback to Activity 6C Have you tried the questions????? If “YES”, check your answers now 6.8 2 740 watt or 2.74 kw 6.9 591 watt. 6.10 1074.5 N ; 8.0 kw ; 17.54 kw.
BELTING J 3010/6/26 SELF-ASSESSMENT 6 You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment 6 given on the next page. If you face any problems, discuss it with your lecturer. Good luck. 1. A pulley is driven by a flat belt running at speed of 600 m/min. The coefficient of friction between the pulley and the belt is 0.3 and the angle lap is 160°. If the maximum tension in the belt is 700 N, find the power transmitted by a belt. 2. A leather belt, 125 mm wide and 6 mm thick, transmits power from a pulley of 750 mm diameter which run at 500 rpm. 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 does not exceed 2.75 MN/m2, find the maximum power that can be transmitted. 3. A flat belt, 8 mm thick and 100 mm wide transmits power between two pulleys, running at 1 600 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 pulleys is 0.3. If the maximum permissible stress in the belt is 2 MN/m2, find (i) Maximum power transmitted; and (ii) Initial tension in the belt. 4. The moment on a pulley, which produces the rotation of the pulley is called: A. Momentum B. Torque C. Work D. Energy 5. If T1 and T2 are the tension in the tight and slack sides of a belt and θ is the angle of contact between the belt and pulley. Coefficient of friction is μ, then the ratio of driving tension will be: T T T T A. 2  e   B. 1  e   C. 1   D. log10 1   T1 T2 T2 T2
BELTING J 3010/6/27 Feedback to Self-Assessment 6 Have you tried the questions????? If “YES”, check your answers now. 1. 3.974 kw ( see example 6.4) 2. 18.97 kw 3. (i) 14.281 kw (ii) 1.3221 kN 4. B. Torque T1 5. B.  e CONGRATULATIONS!!!!….. T2 May success be with you always….

More Related Content

PDF
Spur gear problem and solution
bydodi mulya
 
PDF
Belt conveyor example calculation
byssuser16d252
 
PDF
Theory of machines by rs. khurmi_ solution manual _ chapter 11
byDarawan Wahid
 
PDF
Solution of numerical problem on chimney draught - Part 2 (Notes)
byAVDHESH TYAGI
 
PPTX
Belt drive
byImmanuel alexander
 
PPTX
Dynamic weight transfer in vehicle
byRohan Sahdev
 
DOCX
MOMENT INERTIA OF A FLYWHEEL LAB REPORT
byENGR . GARISON KIPROTICH
 
DOCX
Heat exchanger lab 2
byANDREW MUNYIVA
 
Spur gear problem and solution
bydodi mulya
 
Belt conveyor example calculation
byssuser16d252
 
Theory of machines by rs. khurmi_ solution manual _ chapter 11
byDarawan Wahid
 
Solution of numerical problem on chimney draught - Part 2 (Notes)
byAVDHESH TYAGI
 
Belt drive
byImmanuel alexander
 
Dynamic weight transfer in vehicle
byRohan Sahdev
 
MOMENT INERTIA OF A FLYWHEEL LAB REPORT
byENGR . GARISON KIPROTICH
 
Heat exchanger lab 2
byANDREW MUNYIVA
 

What's hot

PDF
90233240 screw-conveyor-engineering-guide
bythanhson1313
 
PDF
Material, design & analysis of a bicycle frame
byZubair Ahmed
 
PDF
Kiln axis
byRahman Etika Putra
 
PDF
Kiln tire tightening
bymkpq pasha
 
PDF
Belt Drives_2.pdf
byHarikumarA4
 
PPTX
V belt drives
byabdul ahad noohani
 
PDF
Belt drives and chain drives
bydodi mulya
 
PPTX
Belt and Rope drives
byMuhammad Yahya
 
PDF
Gearbox assembly and production drawings
byLahiru Dilshan
 
PPTX
Selection of v belt drive
bySAI SURYA PRAGADA
 
PPTX
Head gear (1)
bySafdar Ali
 
PPTX
Brakes and dynamometer knw
byKiran Wakchaure
 
PPTX
Machine balancing presentation
byHafiz Muhammad Asif
 
PDF
Design of machine elements - Spur gears
byAkram Hossain
 
PDF
Hybrid Rocket propulsion
byDaniel Digre
 
PPTX
Rope drives
byabdul ahad noohani
 
PDF
Fluid Mechanics Chapter 7. Compressible flow
byAddisu Dagne Zegeye
 
PPTX
Belt conveyor
byAlok Mishra
 
DOCX
Pelton wheel experiment
byDickens Mimisa
 
PPT
Se prod thermo_examples_compressor
byVJTI Production
 
90233240 screw-conveyor-engineering-guide
bythanhson1313
 
Material, design & analysis of a bicycle frame
byZubair Ahmed
 
Kiln axis
byRahman Etika Putra
 
Kiln tire tightening
bymkpq pasha
 
Belt Drives_2.pdf
byHarikumarA4
 
V belt drives
byabdul ahad noohani
 
Belt drives and chain drives
bydodi mulya
 
Belt and Rope drives
byMuhammad Yahya
 
Gearbox assembly and production drawings
byLahiru Dilshan
 
Selection of v belt drive
bySAI SURYA PRAGADA
 
Head gear (1)
bySafdar Ali
 
Brakes and dynamometer knw
byKiran Wakchaure
 
Machine balancing presentation
byHafiz Muhammad Asif
 
Design of machine elements - Spur gears
byAkram Hossain
 
Hybrid Rocket propulsion
byDaniel Digre
 
Rope drives
byabdul ahad noohani
 
Fluid Mechanics Chapter 7. Compressible flow
byAddisu Dagne Zegeye
 
Belt conveyor
byAlok Mishra
 
Pelton wheel experiment
byDickens Mimisa
 
Se prod thermo_examples_compressor
byVJTI Production
 

Viewers also liked

PDF
Unit 2b Power Transmission by Belts
byAAR Mahaveer Engineering College
 
PDF
Gear train
byabdul ahad noohani
 
PDF
Mechanical Engineering : Engineering mechanics, THE GATE ACADEMY
byklirantga
 
PDF
Design of machine elements
bybenchmate
 
PDF
Velocity & acceleration diagrams
bysherin_ginige
 
PPTX
Velocity of mechasnism_by_graphical
byaditaya goel
 
PDF
Gears and Gear Trains
byDeepak Sharma
 
PPT
Gear
bySumanth Karanam
 
PDF
7.velocity analysis
byvarun teja G.V.V
 
PPTX
Gear and Gear trains ppt
byujjawalnagda
 
PPTX
Dinesh machine design
byFlt Lt Dinesh Gupta
 
PPT
Electric motor basics
byMahmoud Shaker
 
PPTX
Instantaneous centre
byVinutha Shivalingaswamy
 
PDF
Design_of_Machine_Elements_Spo
byRamil Jay Ureta
 
PDF
Theory of machines
byGIEDEEAM SOLAR and Gajanana Publications, LIC
 
PPT
THE NATURE OF MATERIALS
byLionel Wolberger
 
PPTX
Theory of Machine and Mechanisms (Gears)
byAbhishek Attal
 
PPTX
Study of Gear Technology
bySarath Chandra
 
PDF
Design of machine_elements_
byZainul Abedin
 
PDF
Course Outcome and Program Outcome Calculation(new method)
byRavikumar Tiwari
 
Unit 2b Power Transmission by Belts
byAAR Mahaveer Engineering College
 
Gear train
byabdul ahad noohani
 
Mechanical Engineering : Engineering mechanics, THE GATE ACADEMY
byklirantga
 
Design of machine elements
bybenchmate
 
Velocity & acceleration diagrams
bysherin_ginige
 
Velocity of mechasnism_by_graphical
byaditaya goel
 
Gears and Gear Trains
byDeepak Sharma
 
Gear
bySumanth Karanam
 
7.velocity analysis
byvarun teja G.V.V
 
Gear and Gear trains ppt
byujjawalnagda
 
Dinesh machine design
byFlt Lt Dinesh Gupta
 
Electric motor basics
byMahmoud Shaker
 
Instantaneous centre
byVinutha Shivalingaswamy
 
Design_of_Machine_Elements_Spo
byRamil Jay Ureta
 
Theory of machines
byGIEDEEAM SOLAR and Gajanana Publications, LIC
 
THE NATURE OF MATERIALS
byLionel Wolberger
 
Theory of Machine and Mechanisms (Gears)
byAbhishek Attal
 
Study of Gear Technology
bySarath Chandra
 
Design of machine_elements_
byZainul Abedin
 
Course Outcome and Program Outcome Calculation(new method)
byRavikumar Tiwari
 

Similar to J3010 Unit 6

PDF
Flat belt drives
byMohamed Mohamed El-Sayed
 
DOC
BELTING
byAzizul Izham
 
PDF
Machine design - II - MCQ
byLearn With GeekAlign
 
PDF
Belt proofs.pdf
byMdHarunurRashidMoon
 
PDF
Belt drives extra
byOwusu Attakorah MIET
 
PPTX
belt student1.pptx1234567891234312394132
byahmedmahmoudmahmoud11
 
PPT
DESIGN OF TRANSMISSION SYSTEMS-Belt drive
byChandra Kumar S
 
PDF
Se oct2011
byswapnil shinde
 
PPTX
Belts and Chains 2023 lecture Upload.pptx
byVehlAchari
 
DOCX
Press problems
byVJTI Production
 
PPTX
Unit 3 DESIGN OF POWER TRANSMISSION SYSTEMS.pptx
byDr P DINESHKUMAR
 
PPTX
UNIT IV Belt and chain drives.pptxmmmmmmm
bysachugavhane
 
PPSX
MET 304 Belt drives
byhotman1991
 
PPTX
Types of belt drives.pptx
bysamy709581
 
PPTX
BELT DRIVES _III Design of machine elements -II.pptx
byiqacaitsr
 
PPT
belt_drive.ppt
byMr.C.Dineshbabu
 
DOCX
MET 105 Module 5 power-transmission
byIbrahim AboKhalil
 
PDF
Kinematics Of Machinery Jntu Model Paper{Www.Studentyogi.Com}
byguest3f9c6b
 
DOC
MET 304 Belt drives
byhotman1991
 
PPTX
Lec5.pptx
bySherAli984263
 
Flat belt drives
byMohamed Mohamed El-Sayed
 
BELTING
byAzizul Izham
 
Machine design - II - MCQ
byLearn With GeekAlign
 
Belt proofs.pdf
byMdHarunurRashidMoon
 
Belt drives extra
byOwusu Attakorah MIET
 
belt student1.pptx1234567891234312394132
byahmedmahmoudmahmoud11
 
DESIGN OF TRANSMISSION SYSTEMS-Belt drive
byChandra Kumar S
 
Se oct2011
byswapnil shinde
 
Belts and Chains 2023 lecture Upload.pptx
byVehlAchari
 
Press problems
byVJTI Production
 
Unit 3 DESIGN OF POWER TRANSMISSION SYSTEMS.pptx
byDr P DINESHKUMAR
 
UNIT IV Belt and chain drives.pptxmmmmmmm
bysachugavhane
 
MET 304 Belt drives
byhotman1991
 
Types of belt drives.pptx
bysamy709581
 
BELT DRIVES _III Design of machine elements -II.pptx
byiqacaitsr
 
belt_drive.ppt
byMr.C.Dineshbabu
 
MET 105 Module 5 power-transmission
byIbrahim AboKhalil
 
Kinematics Of Machinery Jntu Model Paper{Www.Studentyogi.Com}
byguest3f9c6b
 
MET 304 Belt drives
byhotman1991
 
Lec5.pptx
bySherAli984263
 

More from mechestud

PDF
J3010 Unit 3
bymechestud
 
PDF
J3010 Unit 5
bymechestud
 
PDF
J3010 Unit 2
bymechestud
 
PDF
J3009 Unit 12
bymechestud
 
PDF
J3009 Unit 5
bymechestud
 
PDF
J3009 Unit 10
bymechestud
 
PDF
J3009 Unit 8
bymechestud
 
PDF
J3009 Unit 11
bymechestud
 
PDF
J3009 Unit 7
bymechestud
 
PDF
J3009 Unit 3
bymechestud
 
PDF
J3009 Unit 6
bymechestud
 
PDF
J3009 Unit 1
bymechestud
 
PDF
J3009 Unit 13
bymechestud
 
PDF
J3010 Unit 4
bymechestud
 
PDF
J3009 Unit 2
bymechestud
 
PDF
J3009 Unit 9
bymechestud
 
PDF
J3010 Unit 0
bymechestud
 
PDF
J3009 Unit 4
bymechestud
 
PDF
J3010 Unit 1
bymechestud
 
PDF
J3009 Unit 0
bymechestud
 
J3010 Unit 3
bymechestud
 
J3010 Unit 5
bymechestud
 
J3010 Unit 2
bymechestud
 
J3009 Unit 12
bymechestud
 
J3009 Unit 5
bymechestud
 
J3009 Unit 10
bymechestud
 
J3009 Unit 8
bymechestud
 
J3009 Unit 11
bymechestud
 
J3009 Unit 7
bymechestud
 
J3009 Unit 3
bymechestud
 
J3009 Unit 6
bymechestud
 
J3009 Unit 1
bymechestud
 
J3009 Unit 13
bymechestud
 
J3010 Unit 4
bymechestud
 
J3009 Unit 2
bymechestud
 
J3009 Unit 9
bymechestud
 
J3010 Unit 0
bymechestud
 
J3009 Unit 4
bymechestud
 
J3010 Unit 1
bymechestud
 
J3009 Unit 0
bymechestud
 

Recently uploaded

PDF
CMOS Fabrication: Step-by-Step Device Processing for ICs — Dr. G. S. Virdi
byGS Virdi
 
PDF
Projects_for_Ministries,_Departments_&_Agencies2123341679.pdf
bynservice241
 
PDF
2025-11-02 - AQSE - Qiskit Fall Fest - IIT-Guwahati.pdf
byAritra Sarkar
 
PDF
Auto Autorefractometer/ppt/notes/pdf.pdf
byAnmol Singh
 
PDF
Unit - I Communication Skills. B.pharmacy and D. Pharmacy
bySayyadTarannum
 
PDF
Gene Pool in Crop Plants: Diversity, Classification, and Role in Plant Breedi...
bySatyam Sharma
 
PDF
MKU Final Arrenged Set.pdf Dhubulia Quiz
byDebanjan Bandyopadhyay
 
PPTX
NURSING PROCESS.....................pptx
byAneetaSharma15
 
PDF
Plant Domestication | Complete Concept Explained BREEDING PLANT DOMESTICATION
bySatyam Sharma
 
PDF
FA Term 2 Reiki Yoga All Student Workshop 22
byLDMMIA, Reiki Yoga
 
PDF
Opthalmoscope ppt-pdf-notesdownload .pdf
byAnmol Singh
 
PDF
Unit4DC_TransportLayer &NetworkLayer.pdf
bySonaShaiju1
 
PPTX
How to Manage Reconciliation of Draft Entries in Odoo 19
byCeline George
 
PPTX
ANKYLOSING SPONDYLITIS OR BAMBOO SPINE.pptx
byDrNidhiAgarwal
 
PDF
Selection in Genetics and Plant Breeding | Complete Concept Explained 🌾
bySatyam Sharma
 
PPTX
Gastro retentive drug delivery system pradip
byDrxpradipPatil
 
PPTX
GROWTH AND DEVELOPMENT..............pptx
byAneetaSharma15
 
PPT
Layers of the Earth crust and Lithosphere.ppt
bybonpat1954
 
PPTX
Presentation1jdhdhhhjjjdbbdnjenmmen.pptx
bypgiopration
 
PDF
Rise of Magadh under Haryanka, Shishunaga and Nanda dynasties
byPrachiSontakke5
 
CMOS Fabrication: Step-by-Step Device Processing for ICs — Dr. G. S. Virdi
byGS Virdi
 
Projects_for_Ministries,_Departments_&_Agencies2123341679.pdf
bynservice241
 
2025-11-02 - AQSE - Qiskit Fall Fest - IIT-Guwahati.pdf
byAritra Sarkar
 
Auto Autorefractometer/ppt/notes/pdf.pdf
byAnmol Singh
 
Unit - I Communication Skills. B.pharmacy and D. Pharmacy
bySayyadTarannum
 
Gene Pool in Crop Plants: Diversity, Classification, and Role in Plant Breedi...
bySatyam Sharma
 
MKU Final Arrenged Set.pdf Dhubulia Quiz
byDebanjan Bandyopadhyay
 
NURSING PROCESS.....................pptx
byAneetaSharma15
 
Plant Domestication | Complete Concept Explained BREEDING PLANT DOMESTICATION
bySatyam Sharma
 
FA Term 2 Reiki Yoga All Student Workshop 22
byLDMMIA, Reiki Yoga
 
Opthalmoscope ppt-pdf-notesdownload .pdf
byAnmol Singh
 
Unit4DC_TransportLayer &NetworkLayer.pdf
bySonaShaiju1
 
How to Manage Reconciliation of Draft Entries in Odoo 19
byCeline George
 
ANKYLOSING SPONDYLITIS OR BAMBOO SPINE.pptx
byDrNidhiAgarwal
 
Selection in Genetics and Plant Breeding | Complete Concept Explained 🌾
bySatyam Sharma
 
Gastro retentive drug delivery system pradip
byDrxpradipPatil
 
GROWTH AND DEVELOPMENT..............pptx
byAneetaSharma15
 
Layers of the Earth crust and Lithosphere.ppt
bybonpat1954
 
Presentation1jdhdhhhjjjdbbdnjenmmen.pptx
bypgiopration
 
Rise of Magadh under Haryanka, Shishunaga and Nanda dynasties
byPrachiSontakke5
 
In this document
Powered by AI

Overview of belting's general and specific objectives, including power transmission concepts.

Conceptual introduction of belt functions in machines and types of belts (flat and V belts).

Detailed mechanics of open belt drive systems with formulas for calculating belt length.

Worked example to compute belt length for specific pulley diameters in an open belt system.

Explanation of closed belt drive systems and formulas for calculating length of the belt.

Worked example for calculating the belt length necessary for a closed belt drive system.

Practical exercises to test understanding of belt length calculation and related concepts.

Formulas and explanation of power transmission via belt tension on different sides.

Exercises to assess knowledge on power and tension calculations for belt applications.

Theoretical derivation of the tension ratio in belts considering friction and mechanics.

Example illustrating how to calculate transmitted power using belt tension and speed.

Discussion on centrifugal tension, its role, and implications on belt performance.

Worked example of torque, tension, and power for an open belt drive system.

Practical activities for users to verify understanding of belt systems and calculations.

Final self-assessment with questions and feedback to reinforce learned belting concepts.

J3010 Unit 6

  • 1.
    BELTING J 3010/6/1 UNIT 6 BELTING OBJECTIVES General Objective : To understand and apply the concept of belting Specific Objectives : At the end of this unit you will be able to:  state the difference between open and close belt.  explain that power transmitted by the flat belt and V belt.  explain that ratio tension for flat and V belt.  calculate power transmitted by the belt with consider centrifugal force.
  • 2.
    BELTING J 3010/6/2 INPUT 6.0 INTRODUCTION In factories, the power has to be transmitted from one shaft to another, then belt driving between pulleys on the shafts may be used. The pulley rotating shaft is called driver. The pulley intended to rotate is known, as follower or driven. When the driver rotates, it carries the belt that grip between its surface and the belt. The belt, in turn, carries the driven pulley which starts rotating. The grip between the pulley and the belt is obtained by friction, which arises from pressure between the belt and the pulley. (a) Types of belts The flat belt is mostly used in the factories and work shops, where a moderate amount of power is to be transmitted, from one pulley to another, when the two pulleys are not more than 10 m a part. The V-belt is mostly used in the factories and work shops where a great amount of power is to be transmitted, from one pulley to another, when the two pulleys are very near to each other.
  • 3.
    BELTING J 3010/6/3 The types of belts:- a. flat belt b. V belt 6.1 LENGTH OF AN OPEN BELT DRIVE A B K α1 α2 C O1 O2 F r2 r1 D E d Fig.6.1 Open belt drive O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = Distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle BO2C = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian)
  • 4.
    BELTING J 3010/6/4 We know that the length of the belt, L = Arc AFE + ED + Arc DCB + BA r1  r2 Cos α1 = d r r α1 = cos-1 1 2 d θ1 = 2π - 2α1 = 2 (π - α1) (radian) θ2 = α1 = 2 α2 (radian) Arc AFE = r1θ1 Arc DCB = r2θ2 And ED = BA = KO2 KO2 Sin α1 = d KO2 = d Sin α1 Finally the total of length of belt, L = Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sin α1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1
  • 5.
    BELTING J 3010/6/5 Example 6.1 Find the length of belt necessary to drive a pulley of 480 cm diameter running parallel at a distance of 12 meter from the driving pulley of diameter 80 cm. This system is an open belt drive. Solution 6.1 A B K F α1 O2 α2 C O1 r2 r1 D E d Fig.6.2 Open belt drive O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = Distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle BO2C = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian) Radius of smaller pulley = 80/2 = 40 cm. Radius of larger pulley = 480/2 = 240 cm. Distance between the pulleys, d = 12m = 1 200 cm We know that the length of belt is, L = Arc AFE + ED + Arc DCB + BA r1  r2 240  40 Cos α1 = = d 1200
  • 6.
    BELTING J 3010/6/6 240  40 α1 = cos-1 1200 = cos-1 0.16667 = 80º α1 = 1.396 radian θ1 = 2π - 2α1 = 2 (π - 1.396 ) (radian) = 3.491 radian θ2 = 2α1 = 2 α2 (radian) = 2 (1.396 ) = 2.792 radian Arc AFE = r1θ1 = 240 x 3.491 = 837.84 cm Arc DCB = r2θ2 = 40 x 2.792 = 111.68 cm And ED = BA = KO2 KO2 Sin 80º = d KO2 = d Sin α1 = 1 200 x 0.98481 = 1181.77 m Finally the total of belt length is L = Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sin α1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1 = 240 x 3.491 + 40 x 2.792 +2 x 1181.77 = 837.84 + 111.68 + 2363.54 = 3313.06 cm = 33.13 m
  • 7.
    BELTING J 3010/6/7 6.2 LENGTH OF CLOSE BELT DRIVE K A D C O1 α1 α2 O2 F r2 r1 B E d Fig.6.3 Cross belt drive O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = Distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle DO2O1 = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian) We know that the length of the belt is L = Arc AFE + ED + Arc DCB + BA r1  r2 Cos α1 = d r r α1 = cos-1 1 2 d θ1 = 2π - 2α1 = 2 (π - α1) (radian) θ1 = θ2
  • 8.
    BELTING J 3010/6/8 Arc AFE = r1θ1 Arc DCB = r2θ2 And ED = BA = KO2 KO2 Sin α1 = d KO2 = d Sin α1 Finally the total of length belt, L = Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sin α1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1 Example 6.2 Find the length of belt for a cross belt drive system. The diameter of the drive pulley is 480 cm which running parallel at a distance of 12 meter from the driving pulley which has a diameter of 80 cm. Solution 6.2 K A D α1 α2 C F O1 O2 r2 r1 B E d Fig.6.4 Cross belt drive
  • 9.
    BELTING J 3010/6/9 O1 and O2 = Centers of two pulleys r1 and r2 = radius of the larger and smaller pulleys d = distance between O1 and O2 L = Total length of the belt. Angle AO1O2 = α1 Angle DO2O1 = α2 Angle AFE = θ1 (radian) Angle BCD = θ2 (radian) Radius of smaller pulley = 80/2 = 40 cm. Radius of larger pulley = 480/2 = 240 cm. Distance between the pulleys, d = 12m = 1 200 cm We know that the length of the belt, L = Arc AFE + ED + Arc DCB + BA r1  r2 240  40 Cos α1 = = d 1200 240  40 α1 = cos-1 1200 -1 = cos 0.23333 = 76.5º α1 = 1.335 radian θ1 = 2π - 2α1 = 2 (π - α1) (radian) = 2 (π - 1.335) (radian) = 3.613 radian θ1 = θ2 Arc AFE = r1θ1 = 240 x 3.613 = 867.12 cm Arc DCB = r2θ2 = 40 x 3.613 = 144.52 cm And ED = BA = KO2 KO2 Sin α1 = d KO2 = d Sin 76.5º =1 200 x 0.9724 ED = BA = KO2 = 1166.88 cm
  • 10.
    BELTING J 3010/6/10 Finally the total of length belt, L= Arc AFE + ED + Arc DCB + BA = r1θ1 + d Sinα1 + r2θ2 + d Sin α1 = r1θ1 + r2θ2 + 2d Sin α1 = 240 x 3.613 + 240 x 3.613 + 2 x 1 200 Sin 76.5º = 867.12 + 144.52 + 2333.76 = 3345.4 cm L = 33.45 m
  • 11.
    BELTING J 3010/6/11 Activity 6A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 6.1 Two pulleys, one with a 450 mm diameter and the other with a 200 mm diameter are on parallel shaft of 1.95 m apart and are connected by a cross belt. Find the length of the belt required and the angle of contact between the belt and each pulley. 6.2 It is required to drive a shaft B at 620 rpm by means of a belt from a parallel shaft A. A pulley of 30 cm diameter on shaft A runs at 240 rpm. Determine the size of pulley on the shaft B. 6.3 Find the length of belt necessary to drive a pulley of 1.4 m diameter running parallel at a distance of 1.7 meter from the driving pulley of diameter 0.5 m. It is connected by an open belt. N1 d 2  N 2 d1 N1 = speed diver in rpm N2 = speed follower in rpm d1 ,d2 = diameter pulley driver and follower.
  • 12.
    BELTING J 3010/6/12 Feedback to Activity 6A Have you tried the questions????? If “YES”, check your answers now 6.1 4.975 m; 199 or 3.473 radian. 6.2. 11.6 cm 6.3 6.51 m
  • 13.
    BELTING J 3010/6/13 INPUT 6.3 POWER TRANSMITTED BY A BELT Power = (T1 –T2) v Where, T1 =Tension in tight side in Newton. T2 = Tension in slack side. V = velocity of belt Fig. 6.5, shows the driving pulley (i.e., driver) A and the follower B. The driving pulley pulls the belt from one side, and delivers the same to the other.The maximum tension in the tight side will be greater than that slack side. Fig. 6.5 Torque exerted on the driving pulley = ( T1 - T2) r1 Torque exerted on the driven or follower = ( T1 - T2) r2 Power transmitted = Force x distance = ( T1 - T2) v
  • 14.
    BELTING J 3010/6/14 Example 6.3 The tension in the two sides of the belt are 100 N and 80 N respectively. If the speed of the belt is 75 meters per second. Find the power transmitted by the belt. Solution 6.3 Given, Tension in tight side, T1 = 100 N Tension in slack side, T2 = 80 N Velocity of belt, v = 75 m/s Let P = Power transmitted by the belt Using the relation, Power = (T1 –T2) v = ( 100 – 80) 75 = 1500 watt = 1.5 kw.
  • 15.
    BELTING J 3010/6/15 Activity 6B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 6.4. Find the tension in the tight side, if the tension in slack side 150 N. The speed of the belt is 58 meters per second and the power transmitted by this belt is 2 kw. 6.5 Diameter of driver is 50 mm. If the driver transmits 8 kw when it is rotating at 300 rev/ min. Calculate velocity of driver. 6.6 The tension in the two sides of the belt are 300 N and 180 N respectively. If the speed of the belt is 50 meters per second, find the initial tension and power transmitted by the belt. 6.7 A 100 mm diameter pulley is belt-driven from a 400 mm diameter pulley. The 400 mm pulley rotates at 480 rev/min. The number of rev/sec of the 100 mm diameter pulley is A. 2 B. 32 C. 120 D. 1920 T1  T2 T0 = 2 Where, T0 = Initial tension in the belt.
  • 16.
    BELTING J 3010/6/16 Feedback to Activity 6B Have you tried the questions????? If “YES”, check your answers now 6.4 115.5 N 6.5 0.785 m/s 6.6 6 000 watt or 6 kw; 240 N 6.7 B. 32
  • 17.
    BELTING J 3010/6/17 INPUT 6.4 RATIO OF TENSIONS. Fig. 6.6 Consider a driven pulley rotating in the clockwise direction as in fig 6.6. Let T1 = Tension in the belt on the tight side. T2 = Tension in the belt on the slack side. ө = An angle of contact in radians (i.e , angle subtended by the arc AB, along which the belt touches the pulley, at the centre) Now consider a small position of the belt PQ, subtending an angle  at the centre of the pulley as shown fig 6.6. The belt PQ is in equilibrium under the following force: i. Tension T in the belt at P. ii. Tension T + T in the belt at Q. iii. Normal reaction R, and iv. Frictional force F =  x R Where  is the coefficient of friction between the belt and pulley. Resolving all the forces horizontally and equating the same,
  • 18.
    BELTING J 3010/6/18   R = ( T + T) sin + T sin (6.1) 2 2 Since the  is very small, therefore, substituting,   sin = in equation 6.1 2 2   R = ( T + T) +T 2 2 T  T  T . = + + 2 2 2 T  = T. ( neglecting ) (6.2) 2 Now resolving the forces vertically,    x R = (T + T) cos - T cos (6.3) 2 2  Since the angle . is very small, therefore substituting cos = 1 in 2 equation 6.3 or, xR = T + T – T = T T R = (6.4)  Equating the values of R from equation 6.2 and 6.4, T Or T. =  T Or = . T Integrating both sides from A to B,  T T1 T T2     0 T1 or loge =  T2 T1 = e  ration of tension for flat belt T2 T1 = e /sin  ration of tension for V belt T2 where  = half angle of groove
  • 19.
    BELTING J 3010/6/19 Example 6.4 Find the power transmitted by a belt running over a pulley of 60 cm diameter at 200 rpm. The coefficient of friction between the pulley is 0.25, angle of lap 160 and maximum tension in the belt is 250 KN. Solution 6.4 Given, Diameter of pulley, d = 60 cm = 0.6 m Speed of pulley N = 200 rpm  dN  x 0.6 x 200 Speed of belt, v = = = 2 = 6.28 m/ sec 60 60 Coefficient of friction, μ = 0.25 160 x  Angle of contact,  = 160 = = 2.7926 radian. 180 Maximum tension in the belt, T1 = 250 kN Let P = power transmitted by the belt. Using the relation, T1 = e  T2 = e 0.25 x 2.7926 T1 = 2.01 T2 T1 250 T2 = = = 124.38 kN 2.01 2.01 Now using the relation, P = (T1 –T2 ) v = ( 250 - 124.38 ) 6.28 = 788.89 watt P = 0.79 kw
  • 20.
    BELTING J 3010/6/20 INPUT 6.5 CENTRIFUGAL TENSION. Fig.6.7 The tension caused by centrifugal force is called centrifugal tension. At lower speeds the centrifugal tension is very small, but at higher speeds its effect is considerable, and thus should be taken into account. Consider a small portion PQ of the belt subtending an angle d at the centre of the pulley as show in fig 6.7. Let M = mass of the belt per unit length, V = linear velocity of the belt, r = radius of the pulley over which the belt runs, Tc = Centrifugal tension acting tangentially at P and Q. Length of the belt PQ, = r d Mass of the belt PQ, M = M r d We know that centrifugal force, = M v2/r
  • 21.
    BELTING J 3010/6/21 Centrifugal force of the belt PQ M x r.d v 2 = r = M x d v 2 The centrifugal tension (Tc) acting tangentially at P and Q keeps the belt in equilibrium. Now resolving the force (i.e, centrifugal force and centrifugal tension) horizontally and equating the same, d 2 Tc sin = M x d v 2 2 since angle d is very small, therefore, substituting d d sin = 2 2 d 2 Tc = M x d v 2 2 Tc = M x d v 2 / d Tc = M v2 T1  Tc  e ratio tension for flat belt. T2  Tc T1  Tc  e   / sin  ratio tension for V belt T2  Tc 6.5.1 Condition for the transmission of maximum power The maximum power, When T c = ⅓T1 It shows that the power transmitted is maximum ⅓ of the maximum tension is absorbed as centrifugal tension. The velocity of belt for maximum transmission of power may be obtained from equation T1 = 3Tc = M v2.
  • 22.
    BELTING J 3010/6/22 3 T1 v2 = M T1 v = 3M Example 6.5 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 0.9 kg/m length and the maximum tension in it is not exceed 2 kN. The larger pulley runs at 200 rev/min. Calculate the torque on each of the two shafts and the power transmitted. Coefficient of friction is 0.3 and angle of lap on the smaller pulley is 168° ( 2.947 radian ). Solution 6.5 Given Diameter of larger pulley, d1 = 1.2 m Radius of the larger pulley, r1 = 0.6 Diameter of smaller pulley, d2 =0.5 m Radius of the smaller pulley, r2 = 0.25 m Distance between two shaft, D = 3.6 m Mass of the belt per meter length, M = 0.9 kg/m Maximum Tension, T1 = 2 kN = 2 000 N Speed of the larger pulley = 200 rpm Velocity of the belt,  x 1.2 x 200 V= = 12.57 m/s 60 T c = Mv2 = 0.9 x 12.572 = 142.2 N
  • 23.
    BELTING J 3010/6/23 e µθ = e 0.3 x 2.947 = 2.421 Using the relation, T1  Tc  e T2  Tc 2000  142.2 = 2.421 T2  142.2 (T2 – 142.2) 2.421 = 1857.8 1857.8 (T2 – 142.2) = = 767.37 2.421 T2 = 767.37 + 142.2 T2 = 909.57 N We know that the torque on the larger pulley shaft (TL), TL = ( T1 – T2) r1 = ( 2000 – 909.57) 0.6 = 654.26 Nm. Torque on the smaller pulley shaft (Ts), Ts = ( T1 – T2) r2 = (2000 – 909.57) 0.25 = 272.61 Nm. Power transmitted, P = ( T1 – T2)v = ( 2000 – 909.57) 12.57 = 13706.71 watt = 13.71 kw
  • 24.
    BELTING J 3010/6/24 Activity 6C TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE TO THE NEXT INPUT…! 6.8 Two pulleys, one with a 450 mm diameter and the other with a 200 mm diameter are on parallel of shafts 1.95 m apart. What power can be transmitted by the belt when the larger pulley rotates at 200 rev/min, if the maximum permissible tension in the belt is 1 kN and the coefficient of friction between the belt and pulley is 0.25. Angle of contact between the belt and larger pulley is 3.477 radian. 6.9 Find the power transmitted by a V drive from the following data: Angle of contact = 84º Pulley groove angle = 45º Coefficient of friction = 0.25 Mass of belt per meter length = 0.472 kg/m Permissible tension = 139 N Velocity of V belt = 12.57 m/s. 6.10 An open belt drive connects two pulleys 1.2 m and 0.6 m, on parallel shafts 3 m apart. The belt has a mass 0f 0.56 kg/m and maximum tension is 1.5 kN. The driver pulley runs at 300 rev/min. Calculate the inertial tension, power transmitted and maximum power. The coefficient of friction between the belt and the pulley surface is 0.3.
  • 25.
    BELTING J 3010/6/25 Feedback to Activity 6C Have you tried the questions????? If “YES”, check your answers now 6.8 2 740 watt or 2.74 kw 6.9 591 watt. 6.10 1074.5 N ; 8.0 kw ; 17.54 kw.
  • 26.
    BELTING J 3010/6/26 SELF-ASSESSMENT 6 You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment 6 given on the next page. If you face any problems, discuss it with your lecturer. Good luck. 1. A pulley is driven by a flat belt running at speed of 600 m/min. The coefficient of friction between the pulley and the belt is 0.3 and the angle lap is 160°. If the maximum tension in the belt is 700 N, find the power transmitted by a belt. 2. A leather belt, 125 mm wide and 6 mm thick, transmits power from a pulley of 750 mm diameter which run at 500 rpm. 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 does not exceed 2.75 MN/m2, find the maximum power that can be transmitted. 3. A flat belt, 8 mm thick and 100 mm wide transmits power between two pulleys, running at 1 600 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 pulleys is 0.3. If the maximum permissible stress in the belt is 2 MN/m2, find (i) Maximum power transmitted; and (ii) Initial tension in the belt. 4. The moment on a pulley, which produces the rotation of the pulley is called: A. Momentum B. Torque C. Work D. Energy 5. If T1 and T2 are the tension in the tight and slack sides of a belt and θ is the angle of contact between the belt and pulley. Coefficient of friction is μ, then the ratio of driving tension will be: T T T T A. 2  e   B. 1  e   C. 1   D. log10 1   T1 T2 T2 T2
  • 27.
    BELTING J 3010/6/27 Feedback to Self-Assessment 6 Have you tried the questions????? If “YES”, check your answers now. 1. 3.974 kw ( see example 6.4) 2. 18.97 kw 3. (i) 14.281 kw (ii) 1.3221 kN 4. B. Torque T1 5. B.  e CONGRATULATIONS!!!!….. T2 May success be with you always….