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- 1. BELT DRIVES
- 2. Functions of belts Transmit motion between shafts that are located at a considerable distance from each other They are not used for exact fixed speed ratio slipping They are very flexible – distance or – the angle between the two shafts.
- 3. Types of Belts
- 4. Belt Pulleys•Flat belts: Crowned pulleys• Round and V- belts grooved pulleys or sheaves• Timing belts toothed wheels or sprockets.
- 5. Layout of Flat belt drive :Non-reversing Open BeltReversing Crossed Belt Reversing Open BeltQuarter Twist Belt drive
- 6. V-Belts – Standard Sizes 1 1 1 2 1 7 4 1 21 8 32 2 29 3 5 13 17 32 4 16 32 32 A B C D E
- 7. Selection of V-Belts Power to be transmitted Speed of the small or large pulley Speed ratio Field of application. Approximate distance between the centres of the two pulleys
- 8. Steps of V-belts Selection: Determine the service factor Based on application, Obtainthe design power from the equation: Design power = transmitted power x service factor Select a suitable belt size from fig.(5. )
- 9. Steps of V-belts Selection: Find the diameter of the small pulley (d) from table (5. ). Find the diameter of the large pulley (D) from the equation: D= d x speed ratio Find the length of the belt using the equation: ( D d )2 L 2C 1.57( D d) 4C
- 10. Steps of V-belts Selection: Obtain the standard length of the belt from table (5. ) Calculate the exact centre distance from the equation b b2 32 ( D d ) C 16 Find the angle of lap (arc of contact), from the equation Angle of lap = 180 ( D d )60 C
- 11. Steps of V-belts Selection: Find the capacity of one belt from the equation: YxS Capacity of one belt = 0.91 XS ZS 3 de The values of X,Y and Z can obtained from table (5. ) The equivalent small pulley diameter de can be obtained from the equation: de = diameter of small pulley x coefficient of small pulley
- 12. Steps of V-belts Selection: Find the power transmitted by one belt from the equation: Power of one belt = belt capacity x length coefficient x coefficient of arc of contact The required number of belts can be obtained from the equation: No. of belts = Design power/ power of one belt
- 13. Example:An engine lathe is driven by a squirrel cage electricmotor through a V-belt. The electric motor runs at1500 rpm with a maximum power of 3 hp. If theinput speed to the engine lathe is 500 rpm and thecentre distance between the motor pulley and thelathe pulley is 30 in. Select a suitable size, lengthand number of belts if the lathe is expected to beworking for two shifts, 16 hours/day.
- 14. Solution: Pitch diameter AC Motor: High torque , High-slip AC Motor: torque, Standard Groove Dimensions SquirrelFrom table (5.1), service factor for machine tools with Squirrel cageSize of belt Minimum Cage Groove ,Synchronous, Split Repulsion-Induction Single-phase Range W D X S ,Series Wound, Slipping, DC Motor E electric motor is = 1.2 recommended Application Phase angle Motor :Shunt ,DC : series wound Compound Wound 2.6 to 5.4wound, 340Engines 0.494:Multi- Engine :Single- cylinder Internal 3/8 A 3 Internal, Combustion 0.490 0.125 5/8Design power = transmitted power x service factor Over 5.4 cylinder 380 0.504 Combustion, Line shafts :ClutchesHour in daily service 4.6to 7.0 340 0.637 B 5.4 3--5 8-10 16-24 0.580 3-5 0.1758-10 ¾ 16-24 ½ = 3 x 1.2Agitators for liquids, Blowers and Over 7.0 380 0.650 exhausts , Centrifugal pumps and7.0 to 7.99 0 = 3.6 hp 1.0 34 1.1 0.879 1.2 1.1 1.2 1.3 compressor , Fan up9.0 10 hp and 8.0 to 12.0 C to 36 0 0.887 0.780 0.200 1 1 11/16 machine tool, Light-duty conveyors 12.0 Over 38 0 0.895From figure (5.3) size A is selectedBelt conveyors for sand, grain, etc.Dough mixers and Fan over 10 hp --12.99 12Then the recommended diameter of small pulley from table (5.2)Generators and line-shafts, Laundry and 13.0 -- 17.0 340 1.259 0 printing machinery 13.0 D , Punches, presses 1.1 36 1.2 1.271 1.3 1.0501.2 0.300 1.3 1 7/16 1.4 7/8 d = 3 in. ,shears , Positive displacement rotary Over 17.0 380 1.283 pumps, Revolving and vibrating screens speed ratio = 1500/500Brick and textile machinery 18.0 to 24.0 360 1.527BucketE elevators and exiters ,Piston 21.0 0 1.300 0.400 1 3/4 1 1/8 Over 24.0 38 1.542 =3 pumps and compressors ,Hammer-mills and paper-mill beaters , Conveyers and 1.2 1.3 1.4 1.4 1.5 1.6Diameter of large pulley (D) = d x speed ratio pulverizers, Positive displacement blowers, Sawmill and wood-working machinery =3x3Crushers ,mills and hoists 1.3 1.4 1.5 1.5 1.6 1.8 = 9 in.Rubber calendars , extruders and mills
- 15. The length of the belt can be obtained from the equation: ( D d )2 L 2C 1.57( D d ) 4C ( 9 3 )2 L 2 x30 1.57( 9 3 ) 4 x30 L = 79.14 in.
- 16. A B C A B C D E Standard Standard Standard Pitch Lengths , Designation Designation Standard Pitch Lengths , Inches InchesFrom table (5.3) the standard length of the … …= 79.3 in 26 27.3 …. ……. belt … 31 32.3 …. …. 107.9 … …with a designation number A78. 33 34.3 …. …. … … 35 36.3 ….. …The exact centre distance from the … 38 42 39.3 43.3 equation: … … ….. … … 46 47.3 b… b2 32(D … 2 d) … 48 49.3 C … … … 51 52.3 16 … … … 53 54.3 … … … 56.3 … … 61.3 b 4 L 6.28( D … d) 63.3 … … 65.3 … … 67.3 b 4x 79.3 6.28(9 3) … … 69.3 … b = 241.84 72.3 76.3 … … … … … 2 … 79.3 … … 241.84 … 241.84 … … … 32(9 3) 2 81.3 …. C … … … …. … 16 … … … 86.3 … … … C = 30.08 in. 91.3 … … … 97.3 … … … … … … …
- 17. The angle of lap (arc of contact), from the equation:Angle of lap = ( D d )60 180 C Angle of lap = ( 9 3 )60 180 30.08 Angle of lap = 168.03o
- 18. Speed Ratio Small Speed Small Speed Small Range Diameter Regular Quality Belts Ratio Diameter Ratio Diameter The capacity of one belt from the equation: Factor Range Factor Belt Cross Section Range Factor 1.341 - 1.429 Capacity of one belt =1.000 - 1.019 FACTORS1.020 - 1.032 1.00 1.01 1.110 - 1.142 BXS 1.05 A 1.143 - 1.178 0.91 YxS ZS 3D - 1.562 E C 1.06 1.430 1.10 1.11 de 1.563 - 1.8141.033 - 1.055 Values of X , 1.179 - 1.222 be Used in H.P. Formula 1.02 Y and Z to 1.07 1.12 1.815 - 2.9481.056 - 1.081 1.03 1.223 - 1.274 1.08 1.13 2.949 - and From table(5.4)1.082 - 1.109 X 1.04 1.945 1.275 - 3.434 1.430 6.372 13.616 19.914 1.14 1.09 over X =2.684 3.801 9.830 26.899 93.899 177.74 Y Z 0.0136 0.0234 0.0416 0.0848 0.1222 Y =5.326 Z =0.0136 Premium Quality Belts S = (3.142 x3 x 1500)/12000 = 1.18 Belt Cross Section de FACTORS =A = 3 x 1.14 3.42 B C D E Values one ,belt = to684Used in91 5.326 x1.18 0.0136 x1.1833 Capacity of of X Y and Z 2. be x1.180. H.P. Formula X 2.684 4.737 8.792 3.42 18.788 24.478 Y 5.326 13.962 38.819 137.70 263.04 Capacity Z of one belt = 1.26 hp 0.0136 0.0234 0.0416 0.0848 0.1222
- 19. Find theStandard Belt Cross Section by one Contact fromCthedrive E Arc of Contacttransmitted Standard of belt Belt Cross Section power A Type ofCdrive Length Length B Arc A B Type of equation: D on V Designation on V to VFactor to Flat V to V Factor to Flat VPower of one belt =belt … Designation Small sheaves 26 Correction 0.81 … capacity x sheaves Correction … … Small length coefficient x … …coefficient of arc of … …Factor 31 33 0.86 … contact 0.84 Correction … Correction Factor … … … … … …Coefficient of arc of contact (from130 35 180 38 0.87 0.88 1.00 … 0.75 table (5.6)= 0.974. 0.86 0.86 …Length coefficient0.95 … 0.77 (5.7)= 1.03. 170 42 160 46 0.90 0.92 0.98 (from table … 0.8 120 110 … … 0.82 0.78 … 0.82 … 0.78 … … … 0.74 …Power of one belt = 1.26 0.82 150 48 51 140 0.93 0.92 0.94 0.89 x 0.974 x901.03… 0.74 0.84 100 … 0.69 0.96 … … … … 53 0.93 = 1.264 hp … … …The required number of…belts can be obtained from the … … …equation: … …No. of belts = Design power/ power…of one belt … … … … …No. of belts = 3.6/ 1.264 … … … … = 2.85…… … … … … …Take 3 belts … … … … … … … … … … … … … … … … … …
- 20. Thank You Dr. Salah Gasim Ahmed MET 103 20

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