Hitachi

Engineering Class Chains Conveyor Chain Selection Selection Examples Example #1: Horizontal Unit Conveyor Supplied Information: Slat Conveyor to move finished TV sets from packaging line to the shipping floor TV set is packaged in an 18" x 18"" x 24" high box Box Weight: 400 lbs each Operating Speed: 125 ft/min Sprocket Centers: 75 ft Infrequent Moderate Shock Loading 16 Hour Operation in Clean Conditions Slats: 1/4" x 24" Long 12 Tooth Sprockets are Specified Step 2: Calculate Conveyor Pull We now use the Conveyor Pull formula for the horizontal conveyor on the preceding pages as follows: P = (2.1Wf+Mf) x C+J* Note J=0 in this case P = [(2.1 x 30.4 x .16) + (266.6 x .16)] x 75 P = (10.2 + 42.7) x 75 = (52.9 x 75) P = 3,968 lbs. Step 3: Calculate Design Conveyor Pull The Design Conveyor Pull (DP) is given by: DP = P x SF x SC x NS DP = 3,968 x 1.32 x 1.05 x 1.02 DP = 6,600 lbs. Step 1: Determine Basic Requirements Step 4: Calculate Chain Tension A unit conveyor in clean operating conditions suggests the use of 2 strands of Roller Conveyor Chain with K-2 attachments to bolt the slats on. The calculated Chain Tension (T) is given by: T = DP / Total Number of Chain Strands T = 6,600 / 2 T = 3,300 lbs. Slat weight will be approximately: Slat Weight=1/4in x 24in x 12in/ft x .2833 lbs/cubic inch Slat Weight = 20.4 lbs/ft of conveyor Material Weight is estimated by: Number of boxes on Conveyor=75ftx12in/ft/18in box=50 Weight of 50 boxes = 50 x 400 = 20,000 lbs Weight per foot of conveyor (M) = 20,000 lbs / 75 ft M = 266.6 lbs/ft of conveyor. To estimate the required chain weight we note the following: Slat weight + Material Weight on 75 ft of conveyor is: (20.4 + 266.6) x 75 = 23,055 lbs. Estimate rolling friction coefficient of .16 from Table 3 we get: 23,055 x .16 / 2 strands = 1,844 lbs of tension per strand. Since 1,844 lbs. is a relatively low tension we estimate the weight on the lower side of what is shown in Table 1. . . We'll estimate 5 lbs/ft. Step 5: Select Chain and Recalculate Chain Tension (T) From the dimensional tables on page 12, 83R has a rated working load of 3,650 lbs, a 2.0" dia. roller and weighs 9.4 lbs/ft with a K-2 every pitch. Recalculate "W". "P". "DP" and "T" as follows: W = (9.6 x 2) + 20.4 = 39.6 P = [(2.1 x 39.6 x .15) + (266.6 x .15)] x 75 P = (12.5 + 40.0) x 75 P = 3,938 lbs. DP = 3,938 x 1.32 x 1.05 x 1.2 DP = 6,550 lbs. T = 6,550 / 2 T = 3,275 lbs. Since the calculated chain tension of 3,275 lbs is less than the rated working load for 83R we can accept this selection. We can now estimate Weight of Chain and Slats per foot of conveyor "W" by the following: Step 6: Verify Roller/Bushing Bearing Pressure W = (5 lbs/ft x 2 strands) + 20.4 = 10 + 20.4 if Chain is Rolling W = 30.4 lbs/ft of conveyor Since we have selected a Roller Conveyor Chain we need to check the roller/bushing bearing pressure as follows: We estimate rolling friction coefficient to be .16 based on a Load Per Roller = (M+W) / [(1/2 / Pitch) x Num. of Strands] lubricated roller. Thus f = .16 Load Per Roller = (266.6 + 39.6) / [(12/4) x 2] Load Per Roller = 306.2 / 6 = 51 lbs. From Table 10 we find the service factor SF by: SF = 1.0 x 1.2 x 1.0 x 1.1 Bearing Area = Bushing OD x (Inside Width - .063) SF = 1.32 Bearing Area = (.875) x (1.313 - .063) = 1.09 Square inches Based on chain speed of 125 ft/min if we use a 12 tooth sprocket we note the speed correction factor: SC = 1.05 From Table 8 since we are using 2 strands of chain we determine the number of strands factor as: NS = 1.2 28 Bearing Pressure = Load Per Roller / Bearing Area Bearing Pressure = 51 / 1.09 = 46.8 psi We refer to the dimensional tables and note that the bushing and roller are both carburized steel. From Table 11 (page 27) allowable bearing pressure is 750 psi. As the calculated bearing pressure of 46.8 is less than 750 psi the selection of 83 is verified.

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