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- 1. UNIT 2 b – Power Transmission by belts By Dr. S. Murali Professor & Head Department of Mechanical Engineering J k h N C ll f E i iJayaprakash Narayan College of Engineering Dharmapur, Mahabubnagar-509001 Telangana IndiaTelangana, India
- 2. Introduction • Rotating elements which possess mechanical h b ili d i d l benergy has to be utilized at required place by transmitting. – From prime mover to machine – From one shaft to anotherFrom one shaft to another 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.2
- 3. Transmission system and Typesy yp • The system that is used to transmit power f h i l l hfrom one mechanical element to another mechanical element • Types Belt drives– Belt drives – Rope drives – Chain drives – Gear drives 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.3
- 4. Factors to select transmission systemy • Distance between driver and driven pulley h fshaft. • Operational speed.p p • Power to be transmitted 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.4
- 5. Belt drives • Power is to be transmitted between the parallel shaftshaft. • Consists of two pulleys over which a endless belt is passed encircling the bothpassed encircling the both. • Rotary motion is transmitted from driving pulley to driven pulleydriven pulley. 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.5
- 6. Terminology of a belt drivegy • Driver : in a transmission system the one which drives or supplies power to other mechanicaldrives or supplies power to other mechanical element. • Driven : in a transmission system the one whichy follows the driver or receives power from driver. • Tight side : the portion of the belt in maximumg p tension. Denoted by T1 Newton. • Slack side : the portion of the belt in minimum tension. Denoted by T2 Newton • Arc / angle of contact : it is the portion of the belt h h h ll f dwhich is in contact with pulley surface. Denoted by 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.6
- 7. Belt material • Rubber • Leather • CanvasCanvas • Cotton • Steel 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.7
- 8. Power transmission systemsy In factories, the power or rotary motion, from h f h id blone shaft to another at a considerable distance is, usually, transmitted by means of belts, ropes and gears. 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.8
- 9. Belt Drives 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.9
- 10. Belt Drives Drilling machine with speed cone pulleys A open belt drive in a jig-saw machine Lathe machine withmachine Lathe machine with speed cones and timing belt 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.10
- 11. Types of Beltsyp (a) Flat Belt (b) V Belt ( ) Ci l B lt(a) Flat Belt (b) V-Belt (c) Circular Belt Moderate amount of power two pulleys Great amount of power two pulleys Great amount of power two pulleyspower two pulleys are not more than 10 m apart. power two pulleys are very nearer to each other. power two pulleys not more than 5 m apart. 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.11 10 m apart. each other. apart.
- 12. Types of Belt drivesyp Open Belt Drive Cross Belt Drive • two pulleys rotate in the same direction • Length of the belt is smaller • pulleys rotate in the opposite directions • Length of the belt is largerLength of the belt is smaller • Angle of lap is different for driver and driven pulley Length of the belt is larger • Angle of lap is same for driver and driven pulley 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.12
- 13. Velocity Ratio(VR)y ( ) It is the ratio between the velocities of the driver and the driven (follower) L d d Di Vb= V1 Let d1,d2 -Diameters, m d1 d2N1 N2 N1, N2 - Speeds, rpm V V Vb -Velocities m/min Length of the belt length covered⎧ ⎫ ⎧ ⎫ For no slip Vb= V2 V1 ,V2 ,Vb -Velocities, m/min Length of the belt length covered passing over driver pulley by a point on driver pulley inone minute in oneminute = ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ Length of the belt length covered⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ 1 1bV d Nπ= 2 2b passing over driven pulley by a point on driven pulley inone minute in oneminute = V d Nπ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ = 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.13 2 2bV d Nπ
- 14. Velocity Ratio(Contd..,)y ( ) Length of the belt Length of the belt passing over driver pulley passing over d= riven pulley ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ 1 1 2 2bV d N d N N dSpeed of Driven inone minute inone minute π π ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ = = 2 1 1 , N dSpeed of Driven Velocity Ratio VR Speed of Driver N = = = 2d 1 N∴ ∝ Vb= V1N d ∴ ∝ From the equation it is obvious that i i l b l d i if di f d1 d2N1 N2 b 1 in simple belt drive if diameter of the pulley decrease Speed will increases and vice versa. Vb= V2 If thickness(t) of the belt is considered 2 1N d t VR N d t + = = + 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.14 1 2N d t+
- 15. Prob: RS Khurmi/Example 33.1/page 672p p g • It is required to drive a shaft at 720 revolutions per minute, by means of a belt from a parallel shaft, having a pulley A 300 mm diameter on it and running at 240 revolutions per minute. What sized pulley is required on the shaft B ? Given: Speed of the driver N = 620 rpm ; Case (2):Given: Speed of the driver, N2 = 620 rpm.; Diameter of pulley,d1 = 300 mm and speed of the pulley, N1 = 240 rpm. Case (2): N2 = 1440 rpm.; d1 = 300 mm and N 240Find Diameter of the follower,d2 Solution: Velocity Ratio VR= 2 1 720 2 58 N d = = = N1 =240 rpm. 2 1 1440 6 N d = = =Velocity Ratio, VR= 2 1 1 2 2.58 240N d = = = 1 2 300d d⇒ = = 2 1 1 2 6 240N d = = = 1 300d d⇒ = =2 3VR 2 100d mm= 2 50d mm= 2 6 d VR ⇒ 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.15 2 2
- 16. Limitations of simple belt drivep • Higher Velocity ratios ratios, lessen the arc of i li d l fcontact, causing slippage and loss of power. • For maximum power transfer on the beltsp and pulleys, the pulley ratio should be 3 to 1 or lessor less • To avoid excessive single-step ratios or undersize pulleys. – The central distance can be increased – Compound drive /two-step drive /counter-shaft) can be used tocan be used to 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.16
- 17. Compound belt drivep • power is transmitted, from one shaft to another, through a number of pulleysnumber of pulleys • Consists of more number of simple belt drivesp • Pulleys 2 and 3 are keyed to the same shat • Pulley 1 and 2 forms one belt drive and Pulley 3 and 4 another 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.17
- 18. Compound belt drivep Let d1,d2, d3, d4, and N1,N2, N3, N4= diameters and speeds for pulleys 1, 2, 3 and 4. N d We know that velocity ratio of pulleys 1 and 2, 2 1 1 1 2 N d VR N d = = Similarly, velocity ratio of pulleys 3 and 4, ---------------- (1) y, y p y , 34 2 3 4 dN VR N d = = ---------------- (2) Multiplying the above equations 1 32 4 1 2 d dN N VR VR N N d d ×× × = = × × ---------------- (3) 1 3 2 4N N d d× × Since N2=N3, therefore velocity ratio of compound belt drive 1 34 1 2 d dN VR VR VR N d d × = × = = × ---------------- (4) 1 2 4N d d× Speed of last follower product of diametersof drivers VR Speed of first driver product of diametersof followers = = 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.18 Speed of first driver product of diametersof followers
- 19. Problem: It is required to drive a shaft at 1440 revolutions per minute, by means of compound belt from a parallel shaft, having drivers 300 mm diameter and first driver is running t 240 l ti i t Wh t i d ll i d f f ll ?at 240 revolutions per minute. What sized pulley are required for followers ? Given: Speed of the driver, N4 = 620 rpm.;4 Diameter of pulleys,d1 = d3 =300 mm and speed of the pulley, N1 = 240 rpm. Find Diameter of the follower d and dFind Diameter of the follower,d2 and d4 Solution: Velocity Ratio compound drive, For Case 1 1 34 1 2 1 2 4 1440 6 240 d dN VR VR VR N d d × = × = = = = × 1 2 1 300 150 2 d d mm VR = = = 1 2 6 2 3 1 3 2 2 VR VR VR Case Case ×⎧ ⎪ = × ←⎨ ⎪ × ←⎩ 3 4 2 300 100 3 d d mm VR = = = 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.19 3 2 2Case⎪ × ←⎩
- 20. Slipp • If there is no firm frictional grip between the belts and the shafts. This may cause forward motion of the driver pulley without carrying the belt with it or forward motion of the belt without carrying the driven pulley with it. This is called slip of the belt, and is generally expressed as a percentage and denoted by S. V V Vb V1 Slip at driver,S1,Expressed in % of driver speed, 1 1 1 % 100bV V V S − = × d1 d2N N2 2N1 V Vb V2 Slip at driven,S2,Expressed in % of belt speed 2 2 % 100bV V V S − = × 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.20 2 bV
- 21. Slipp 1 1 1 100bV V V S − = × ⎛ ⎞ N l ti 1 2S S× 1 1 1 100 bV V S⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ ---------------- (1) V V Neglecting 12 1 2 1 100 d d SN N S+⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ 1 2 10000 S S× ---------------- (2) 2 2 2 2 % 100 1 b b b V S V V V S V − = × ⎛ ⎞ = −⎜ ⎟ 1 2 100dN ⎜ ⎟ ⎝ ⎠ Assuming S1+S2=S N d S⎛ ⎞ (2)2 1 100 bV V ⎜ ⎟ ⎝ ⎠ S b tit ti th l V i (2) 2 1 1 2 1 100 N d N S d ⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ Substituting the value Vb in eq (2) 1 2 1 2 1 1 100 100 S S V V ⎛ ⎞⎛ ⎞ = − −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ If thickness(t) of the belt is considered 2 1 1 2 1 100 N d t N d t S+ ⎛ ⎞ = −⎜ ⎟ + ⎝ ⎠ 2 2 1 1 1 2 1 2 1 100 100 10000 S S N d S N S dπ π ⎛ ⎞ = × − − +⎜ ⎟ ⎝ ⎠ 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.21
- 22. Length of an Open Belt Driveg p From geometry O1M= O1E- ME= r1- r2 From right angled triangle O1MO2g g g 1 2 1 2 sin r r x α − = For small values of α, sinα=α 1 2r r− (1)1 2 x α = Arc length JE= 1 2 r π α ⎛ ⎞ +⎜ ⎟ ⎝ ⎠ ---------------- (1) ---------------- (2)g 2 ⎜ ⎟ ⎝ ⎠ (∵ l=rθ ) (2) Arc length FK= 2 2 r π α ⎛ ⎞ −⎜ ⎟ ⎝ ⎠ ---------------- (3) 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.22
- 23. Length of an Open Belt Driveg p ( ) 22 2 1 2MO EF x r r= = − − (pythagoras theorem) 2 r r⎛ ⎞1 2 1 r r x x −⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ 2 4 1 2 1 21 1 1 r r r r x ⎛ ⎞− −⎛ ⎞ ⎛ ⎞ = − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ (Binomial expansion) 2 1 21 1 r r⎛ ⎞−⎛ ⎞ ⎜ ⎟⎜ ⎟ (neglecting higher order terms) 1 2 8 x x x +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ( p ) ( ) 2 1 2 2 1 2 r r MO EF x − = = − 1 2 1 2 x x = −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ (neglecting higher order terms) ---------------- (4)2 2 x Length of the belt =Length of arc GJE + EF + Length of arc FKH + HG L=2 (Length of arc JE + EF + Length of arc FK) 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.23
- 24. Length of an Open Belt Driveg p Substituting the values of length of arc JE from equation (2), length of arc FK from equation (3) and EF from equation (4) in this equation,q ( ) q ( ) q , ( ) 2 1 2 1 2 1 2 2 2 2 r r L r x r x π π α α ⎡ ⎤⎛ ⎞−⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟= + + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦ ( ) ( ) ( ) 2 1 2 2 2 r r r r x r rπ α − + + +( ) ( )1 2 1 22 2r r x r r x π α= + + − + − ( ) ( ) ( ) ( ) 2 1 2 1 2r r r r− − ( ) ( ) 2 1 2 2 r r L + ( ) ( ) ( ) ( )1 2 1 2 1 2 1 22 2r r x r r x x π= + + − + − ( ) ( )1 2 1 2 2L r r x x π= + + + 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.24
- 25. Length of an Cross Belt Driveg From geometry O1M= O1E- ME= r1+r2 From right angled triangle O1MO2g g g 1 2 1 2 sin r r x α + = For small values of α, sinα=α 1 2r r α − = ---------------- (1) x Arc length JE= 1 2 r π α ⎛ ⎞ +⎜ ⎟ ⎝ ⎠ (1) ---------------- (2)2 ⎜ ⎟ ⎝ ⎠ (∵ l=rθ ) (2) Arc length FK= 2 2 r π α ⎛ ⎞ +⎜ ⎟ ⎝ ⎠ ---------------- (3) 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.25
- 26. Length of an Cross Belt Driveg ( ) 22 2 1 2MO EF x r r= = − + (pythagoras theorem) 2 r r+⎛ ⎞1 2 1 r r x x +⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ 2 4 1 2 1 21 1 1 r r r r x ⎛ ⎞+ +⎛ ⎞ ⎛ ⎞ = − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ (Binomial expansion) 2 1 21 1 r r⎛ ⎞+⎛ ⎞ ⎜ ⎟⎜ ⎟ (neglecting higher order terms) 1 2 8 x x x +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ( p ) ( ) 2 1 2 2 1 2 r r MO EF x + = = − 1 2 1 2 x x = −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ (neglecting higher order terms) ---------------- (4)2 2 x L th f th b lt L th f GJE EFLength of the belt =Length of arc GJE + EF + Length of arc FKH + HG L=2 (Length of arc JE + EF + Length of arc FK) 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.26
- 27. Length of an Cross Belt Driveg Substituting the values of length of arc JE from equation (2), length of arc FK from equation (3) and EF from equation (4) in this equation,q ( ) q ( ) q , ( ) 2 1 2 1 2 1 2 2 2 2 r r L r x r x π π α α ⎡ ⎤⎛ ⎞+⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟= + + − + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦ ( ) ( ) ( ) 2 1 2 2 2 r r r r x r rπ α + + + + +( ) ( )1 2 1 22 2r r x r r x π α= + + − + + ( ) ( ) ( ) ( ) 2 1 2 1 2r r r r+ + ( ) ( ) 2 1 2 2 r r L + ( ) ( ) ( ) ( )1 2 1 2 1 2 1 22 2r r x r r x x π= + + − + + ( ) ( )1 2 1 2 2L r r x x π= + + + 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.27
- 28. Prob:RS Khurmi/Example 33.4/679p Find the length of belt necessary to drive a pulley of 500 mm diameter running parallel at a distance of 12 meters from the driving pulley of diameter 1600 mm. Given: Diameter of the driven pulley (d2) = 500 mm = 0·5 m or radius (r2) = 0·25 m; Diameter of the driving pulley (d ) 1600 mm 1 6 m or radius (r ) 0 8 mDiameter of the driving pulley (d1) = 1600 mm = 1·6 m or radius (r1) = 0·8 m. Distance between the centres of the two pulleys (x) = 12 m Find length of the belt for 1. open belt drive 2. Cross belt driveFind length of the belt for . open belt drive . Cross belt drive Solution: 1. Open Belt drive ( ) ( ) 2 2 0 8 0 25 ( ) ( ) ( ) ( )1 2 1 2 0.8 0.25 2 0.8 0.25 2 12 27.32 r r L r r x m x x π π − − = + + + = + + × + = 2 C B lt d i2. Cross Belt drive ( ) ( ) ( ) ( ) 2 2 1 2 1 2 0.8 0.25 2 0.8 0.25 2 12 27.39 r r L r r x m x x π π + + = + + + = + + × + = 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.28 x x
- 29. Open belt drive Vs Close belt drive Item Open Belt Drive Closed Belt Drive Arrangement Rotation of pulleys same direction opposite direction useful alignment of shafts horizontal or inclined horizontal or inclined or verticalvertical rubbing no rubbing point, the life of the belt is more rubbing point, the life of the belt reduces. L th f th b ltLength of the belt (same centre distance, pulley diameters.) less length Require more length A l fAngle of contact (big and small pullys) different Same 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.29
- 30. Power transmitted by a belt • the effective turning (i.e. driving) force at the circumference of the follower is F T T y F=T1 – T2. • Work done per second = Force × Distance = T1 – T2 × v ( ) /Power = (T1 – T2)v J/s 2 1 1 1 T P T T ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 1 1 11 2 1 1 1 P T v T v TT T ⎜ ⎟ = − = −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ 1 1 1P T v eμθ ⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ Notes: 1. The torque exerted on the driving pulley =(T1 – T2) r1 2. Similarly, the torque exerted on the follower = (T1 – T2) r2 where r1 and r2 are in metres. 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.30 1 2
- 31. Centrifugal tensiong • The belt continuously runs over both the pulleys. It carries some centrifugal force in the belt atIt carries some centrifugal force in the belt, at both the pulleys, • whose effect is to increase the tension on both• whose effect is to increase the tension on both, tight as well as the slack sides. Th t i d b th t if l f i• The tension caused by the centrifugal force is called centrifugal tension. • At lower speeds, the centrifugal tension is very small and may be neglected. • But at higher speeds, its effect is considerable, and thus should be taken into account. 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.31
- 32. Centrifugal tensiong m = Mass of the belt per unit length, v = Linear velocity of the belt,y , r = Radius of the pulley over which the belt runs, TC = Centrifugal tension acting tangentially at P and Q and dθ = Angle subtended by the belt AB at the centre of the pulley. Length of belt AB = r dθ l f h b l dθtotal mass of the belt M = mr dθ 2 2 C Mv mrd v P r r θ = =We know that centrifugal force of the belt AB, 2 Now resolving the forces (i.e., centrifugal force and centrifugal tension) horizontally and equating the same 2 CP md vθ= and equating the same, 2 2 sin 2 C d T md v θ θ= 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.32
- 33. Centrifugal tensiong For small values of ,sin 2 2 2 d d dθ θ θ = 2 2 sin 2 C d T md v θ θ= 2 TCT mv= Notes: 1. When the centrifugal tension is taken into account, the total tension in the tight side = T1 + TC and total tension in the slack side = T2 + TC 2 Th t if l t i th b lt h ff t th2. The centrifugal tension on the belt has no effect on the power transmitted by it. The reason for the same is that while calculating the power transmitted, we have to use the values : Total tension in tight side Total tension in the slack side= Total tension in tight side – Total tension in the slack side = (T1 + TC) – (T2 + TC) = (T1 – T2). 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.33
- 34. Maximum tension in the belt • Let σ = Maximum safe stress in the belt, N/mm2 b = Width of the belt in mm andb = Width of the belt in mm, and t = Thickness of the belt in mm. • We know that maximum tension in the belt• We know that maximum tension in the belt, T = Maximum stress × Cross-sectional area of beltbelt = σ bt Wh t if l t i i l t d th• When centrifugal tension is neglected, then maximum tension, T = T1 and • when centrifugal tension is considered then• when centrifugal tension is considered, then maximum tension, T = T1 + TC 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.34
- 35. Condition for transmission of maximum power • that the power transmitted by a belt, 1 1 1 1 P T v eμθ ⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ ⎛ ⎞⎛ ⎞ 1 1 assuming 1P TCv C eμθ ⎛ ⎞⎛ ⎞ = = −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ B t T T T and T m 2But T1=T-Tc and Tc=mv2 ∴P= (Tv-mv3)C Differentiating w.r.t v and equating to zero ( )3dP d Tv mv= − 3T T=( ) 2 2 0 3 Tv mv dv dv T mv= − 3 cT T= It shows that when the power transmitted is maximum 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.35 2 3T mv= 1/3 rd of the maximum tension is absorbed as centrifugal tension
- 36. Belt speed for maximum powerp p for maximum power transmission, 2 3T mv= Speed of the belt for maximum transmission of power 3 T v m = 3m Note: Maximum tension in the belt is equal to sum of tensions in tight side (T1) and centrifugal tension (TC). 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.36
- 37. Initial tension in the belt • the motion of the belt (from the driver) and the follower (from the belt) is governed by a firm grip due to friction(from the belt) is governed by a firm grip due to friction between the belt and the pulleys, therefore the belt is tightened up, in order to keep a proper grip of the belt over the pulleys. Initially, even when the pulleys are stationary the belt is subject to some tension, called initial tension. Let T0 = Initial tension in the belt, T1 = Tension in the tight side of the belt, T2 = Tension in the slack side of the belt, and n = Coefficient of increase of the belt length per unit force. 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.37
- 38. Initial tension in the belt that increase of tension in the tight side = T1 – T0 increase in the length of the belt on the tight sideincrease in the length of the belt on the tight side = n (T1 – T0) Similarly, decrease in tension in the slack side = T0 – T2Similarly, decrease in tension in the slack side T0 T2 decrease in the length of the belt on the slack side = n (T0 – T2)( 0 2) For constant length of the belt, when it is at rest or in motion, The increase in length on the tight side = equal to decrease in the length on the slack side Therefore, equating (T T ) (T T ) Note: If centrifugal tension is takenn (T1 – T0) = n (T0 – T2) ∴ T=(T1+T2)/2 Note: If centrifugal tension is taken into consideration, then T=(T1+T2)/2+Tc 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.38
- 39. References 1. R.S. Khurmi & J.K . gupta,2005, “a text book f hi d i ” fi l i l di iof machine design”, first multicolor edition, eurasia publishing house (pvt.) Ltd. 2. K.L. Narayana, P. Kannaiah & K. Venkat reddy 2006 ”machine drawing” 3rd editionreddy,2006, machine drawing , 3 edition, new age international (p) limited, publishers 3. wikipedia.org 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.39
- 40. Contact: smuralichinna@gmail.com 10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.40
- 41. THE BINOMIAL THEOREM (a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 + nC3 an-3 x3 n + nC4 an-4 x4 + ... + nCn xn = kkn n k k n xaC − = ∑ 0 nk 0 (1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + ... + nCn xn = k n k k n xC∑ =0!n n ( ) ( ) ! ! ! ! ! 0! 1 1 n k n n C n k k n n C = − = = = =∵ ( ) ( ) ( ) ( ) ( )1 0! 1 1 0 !0! ! 1 1 !! 1 !1! 1 ! o n C n n n nn C n = = = = − × × − = = = ∵ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 !1! 1 ! 1 2 ! 1! 2 !1! 2 ! 2! 2! n n n n n n n nn C n n − − × − × − × − = = = − − ×( ) ( )2 !1! 2 ! 2! 2!n n × 1010--0202--20152015 1.1.4141Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCEDr. S. Muali, Prof & Head, Department of Mechanical, JPNCE
- 42. Pythagoras theorem (Sulba Sutra) "In any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs" relationship can be stated as: Pythagoras of Samos BAUDHAYANA(800BCE) Pythagoras of Samos (Greek: . 570 BC – c. 495BC) Ionian Greek philosopher and mathematician U (800 C ) “dīrghasyākṣaṇayā rajjuH pārśvamānī, tiryaDaM mānī, cha yatpṛthagbhUteg y ṣ ṇ y jj p , y , y pṛ g kurutastadubhayāṅ karoti.” There are certain sets of numbers that have a very special property. Not only do these numbers satisfy the Pythagorean Theorem, but any multiples of these numbers also satisfyfy y g , y p f fy the Pythagorean Theorem (numbers 3, 4, and 5) 1010--0202--20152015 1.1.4242Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCEDr. S. Muali, Prof & Head, Department of Mechanical, JPNCE

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