Machinery's Handbook, 31st Edition
2586 Double V-Belts 8) In most drives, slippage will occur first at the driver sheave. Assume this to be true and calculate T T and T S for the driver: T T = T e [ R /( R − 1)] and T s = T T − T e . Use R /( R − 1) from Step 7 and T e from Step 6 for the driver sheave. 9) Starting with the first driven sheave, determine T T and T S for each segment of the drive. The T T for the driver becomes T S for that sheave and is equal to T T − T e . Proceed around the drive in like manner. 10) Calculate actual R /( R − 1) for each sheave using: R /( R − 1) = T T / T e = T T /( T T − T S ). The T T and T S values are for those determined in Step 9. If these values are equal to or greater than those determined in Step 7, the assumption that slippage will first occur at the driver is correct and the next two steps are not necessary. If the value is less, the assumption was not correct, so proceed with Step 11. 11) Take the sheave where the actual value R /( R − 1) (Step 10) is less than the minimum, as determined in Step 7, and calculate a new T T and T S for this sheave using the minimum R /( R − 1) as determined in Step 7: T T = T e [ R /( R − 1)] and T S = T T − T e . 12) Start with this sheave and recalculate the tension in each segment of the drive as in Step 9. 13) The length-flex factor ( K f ) is taken from Table 20. Before using this table, calculate the value of L e / n . Be sure to use the appropriate belt cross section column when selecting the correction factor. 14) Beginning with the driver sheave, determine the number of belts ( N b ) needed to sat isfy the conditions at each loaded sheave using: N b = T T / T r K f . Note: T T is tight side tension as determined in Step 9 or 11 and 12. T r is allowable tight side tension as shown in Table 18-Table 21. K f is the length-flex correction factor from Table 20. The sheave that requires the largest number of belts is the number of belts required for the drive. Any fraction of a belt should be treated as a whole belt. Table 16. Allowable Tight Side Tension for an AA Section Belt Speed (fpm) Sheave Effective Diameter (in.) 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 200 30 46 57 66 73 79 83 88 400 23 38 49 58 65 71 76 80 600 18 33 44 53 60 66 71 75 800 14 30 41 50 57 63 67 72 1000 12 27 38 47 54 60 65 69 1200 9 24 36 45 52 57 62 66 1400 7 22 34 42 49 55 60 64 1600 5 20 32 40 47 53 58 62 1800 3 18 30 38 46 51 56 60 2000 1 16 28 37 44 50 54 58 2200 … 15 26 35 42 48 53 57 2400 … 13 24 33 40 46 51 55 2600 … 11 23 31 39 44 49 53 2800 … 9 21 30 37 43 47 51 3000 … 8 19 28 35 41 46 50 3200 … 6 17 26 33 39 44 48 3400 … 4 16 24 31 37 42 46 3600 … 2 14 23 30 35 40 44 3800 … 1 12 21 28 34 38 43 4000 … … 10 19 26 32 37 41 4200 … … 8 17 24 30 35 39 4400 … … 6 15 22 28 33 37 4600 … … 4 13 20 26 31 35 4800 … … 2 11 18 24 29 33 5000 … … … 9 16 22 27 31 5200 … … … 7 14 20 24 28 5400 … … … 4 12 17 22 26 5600 … … … 2 9 15 20 24 5800 … … … … 7 13 18 22 The allowable tight side tension must be evaluated for each sheave in the system (see Step 14). Values must be corrected by K f from Table 20.
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