4. How to Select Chains
Step 5. Confirm that you can set a 15-tooth sprocket and a 38-tooth sprocket within the 1,500-mm center distance and still maintain clearance. The maximum hub bores of each sprocket are 89 and 110, respectively. Therefore, these may be used. Step 6. Calculate L, the number of chain pitches.
center distance chain pitch
C =
1,500 44.45
C =
= 33.746 (sprocket center distance, in pitches)
( N - N' ) ( —— ) 2 6.28
(38 - 15) ( —— ) 2
( N + N' ) 6.28 L = ——-— + 2C + —--—-- = ———- + 2 3 33.746 + ——— 2 C 2 33.746 = 94.39 links (38 + 15) Because you can’t have fractions of links, choose the next highest even number. In this example, you would use 96 pitches. The center distance of the sprockets will then be 1,536 mm. Step 7. Check the catalog and decide the appropriate type of lubrication (manual or drip). 4.1.4 Power Transmission Chain Selection for Slow Speeds This selection procedure is based on the maximum allowable tension, which is used when the chain speed is less than 50 m/min., and the starting frequen- cy is less than 5 times/day. The selection is done following the flow chart in Figure 4.11. EXAMPLE: Recalculate the previous example from Basics Section 4.1.3 based upon the selection for slow speed. Step 1. Tentatively select RS120 chain, which is one size smaller than RS140, and a 15-tooth sprocket. Then calculate the chain speed. V = PNn / 1,000 = (38.1 3 15 3 50)/ 1,000 = 28.6 m/min. < 50 According to this speed and starting frequency, case selection for slow speed may be used. Step 2. From the rated power of the motor, calculate the tension Fm on the chain. Fm = 60 3 kW / V = 60 3 7.5 / 28.6 = 15.7 kN Step 3. Service factor Ks = 1.3, Chain speed coefficient Kv = 1.06 (from the chain speed 28.6 m/min.). Step 4. Sprocket tooth coefficient Kc = 1.27 (from 15-tooth sprocket). Step 5. Calculate the design chain tension F'm. F'm = Fm 3 1.3 3 1.06 3 1.27 = 27.5 kN Step 6. Decide on the chain size.
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