4.1.3 Drive Chain Selection (General Selection)

A suitable chain selection may be made according to the flow chart Figure 4.8. EXAMPLE: Select a transmission chain for the conditions shown in Figure 4.9.

Figure 4.9 Operating Conditions for Example 4.1.3

Type of Application: Drive of Belt Conveyor
Source of Power: Electric Motor 7.5 kW
Drive Shaft: Diameter 66 mm. 50 rpm
Driven Shaft: Diameter 94 mm. 20 rpm
Center Distance of Shafts: 1,500 mm
Starting Frequency: 4 times/day
Type of Impact: Some Impact
Reducer Ratio: 1/30

Step 1. Confirm the operating conditions.
Step 2. Determine the service factor Ks as shown in Table 4.2. In this example, the service factor is Ks = 1.3.
Step 3. Calculate the corrected design power kW = 1.3 3 7.5 = 9.75 kW.
Step 4. Consult the selection table (Figure 4.10). For n = 50 rpm and corrected power = 9.75 kW, you should initially select RS140 chain and a 15-tooth drive sprocket. These are not the final selections. See manufacturer's catalog for additional information.

Figure 4.8 Chain Selection Procedure (General Selection)

Figure 4.10 RS Roller Chain Provisional Selection Table

Step 4a. Calculate the size of the driven sprocket. Number of teeth in driven sprocket = 15 3 (50/20) = 37.5. Therefore, select a 38-tooth driven sprocket.
Step 4b. Confirm that the chain meets the power requirements. According to the power transmission tables in the catalog, an RS140 chain with a 15-tooth sprocket is capable of transmitting 11.3 kW. Because 11.3 kW is greater than the design power of 9.75 kW, it is acceptable.
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.
C = center distance ÷ chain pitch

C = 1,500 ÷ 44.45 = 33.746 (sprocket center distance, in pitches)

L = (N + N′) / 2 + 2C + ((N − N′) / 6.28)²/ C = (38 + 15) / 2 + 2 × 33.746 + ((38 − 15) / 6.28)² / 33.746

= 94.39 links
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).