Design Data - Step 4: Chain Pull Calculation
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Design Data
Step 1: Assemble Job Information
Step 2: Conveyor Component Data
Step 3: System Design
Step 4: Chain Pull Calculation
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Step 4: Chain Pull Calculation

WHAT IS CHAIN PULL?

Chain pull is the effort necessary to maintain the normal operating speed of a conveyor under a related load. To arrive at this figure, it is necessary to add the lift load and the friction factors, expressed as a small percentage of the live load, which act as resistance to the progress of the conveyor. Friction resistance is found in the bearings of the chain wheels, curves, and the drive unit itself. It should be noted that these percentages are for average conveyors that travel under normal conditions. When adverse environment conditions exist or the conveyor is abnormally long or complex and exceeds the chain pull capacity of one drive, a progressive chain pull computation is necessary where the friction losses are progressively calculated and accumulated through the path along the conveyor.

There are few, if any, short cuts in the process of determining the chain pull on an endless chain conveyor. Many factors, which vary greatly from system to system, influence the end result. Product carrier weight, product weight, coefficient of friction, track curves, inclines, declines, temperature and lubrication are some of the principle items affecting chain pull. However, by knowing the value of the various factors, the total chain pull may be determined by the point to point accumulation method of calculation.

ZIG-ZAG FRICTION FACTORS

ITEM

%

Straight

1.50%

45° Vertical Curves

1.00%

90° Vertical Curves

2.00%

180° Vertical Curves

4.00%

45° Horizontal Curves

1.00%

90° Horizontal Curves

2.00%

180° Horizontal Curves

4.00%

540° Ransburg Horizontal

12.00%

90° or 180° Wheel Turn

3.00%

The above friction factors were obtained from accurate laboratory tests with lubricated chain and have been verified by dynamometer tests of complete systems.

It will be seen that straight conveyor has a friction factor of 1.50%. Therefore, if a conveyor did not have any vertical or horizontal curves, inclines or declines, the weight of the total live or moving load could be multiplied by 1.50% to determine the weight in pounds to move the conveyor. For instance, a 100'-0" conveyor with 1.5 lb. hooks and 6 lb. loads on 1'-0" centers would be 10 lbs. per foot (including chain @ 2.5 lbs. per foot) x 100'-0" or 1000 lbs. @ 1.50% = 15 lbs. chain pull. However, an endless conveyor chain must at least have horizontal curves, and each of these generate additional forces which must be taken into account as will be seen in Figure #1.

CHAIN PULL CALCULATION FOR A LEVEL SYSTEM

In Figure #1 a simple system about 115'-0" long is shown with four horizontal curves. The first item to be determined is the pounds in chain pull per foot of straight conveyor. The following formula provides a result of .45 pounds per foot.

Figure #1

3 lb. Chain

+

2 lb. Pendant

+

6 lb. Hook

+

47 lb. Load

1 ft.

2 ft.

2 ft.

2 ft.

= 30.5 lbs. per foot x 1.50% = .4575 lbs./ft. chain pull

In the following computation, it will be seen that the footage from the down stream end of the drive to the far tangent of the next curve (0 to 1) is multiplied by .45 lbs. and the 2% friction for the 90° curve is added. Next, the footage from (1) to (2) is multiplied by .45 lbs., added to the 3.08 lb. resultant plus 2% for the 90° curve. The procedure of accumulating at each curve is followed back to "0".

0 to 1 = 6.7 ft. x .4575 lbs./ft. + 2% =

3.13 lbs.

1 to 2 = 8.1 ft. x 4575 lbs./ft. + 3.13 lbs. + 2% =

6.97 lbs.

2 to 3 = 53.1 ft. x .4575 lbs./ft. + 6.97 lbs. + 2% =

31.89 lbs.

3 to 4 = 8.1 ft. x .4575 lbs./ft. + 31.89 lbs. + 2% =

36.31 lbs.

4 to 0 = 42.4 ft. x .4575 lbs./ft. + 36.31 lbs. =

55.71 lbs.

TOTAL CHAIN PULL =

55.71 lbs.

CHAIN PULL CALCULATION FOR A MULTI-LEVEL SYSTEM

Introduction of inclines and declines in a system further complicates chain pull calculation by adding to the forces. In Figure #2, a conveyor is shown identical to Figure #1 except for the addition of an incline and decline. Due to the elevations changes, the formula for finding the pounds per foot chain pull is arranged differently because it must be assumed that at some point the incline will be loaded and the decline empty. In systems having multiple rises and falls, this may be treated somewhat differently, depending on loading patterns.

Figure #2

0 to 1 = 11 ft. x .4575 lbs./ft. - (6'-0” x 30 lbs./ft.) =

0 lbs.

1 to 2 = 6.7 ft. x .4575 lbs./ft. + 0 lbs. + 2% =

3.13 lbs.

2 to 3 = 8.1 ft. x .4575 lbs./ft. + 3.13 lbs. + 2%

6.97 lbs.

3 to 4 = 13.6 ft. x .4575 lbs./ft. + 6.97 lbs. + (6'-0” x 30 lbs./ft.) + 2% =

197.06 lbs.

4 to 5 = 37.9 ft. x .4575 lbs./ft. + 197.06 lbs. + 2% =

218.69 lbs.

5 to 6 = 8.1 ft. x .4575 lbs./ft. + 218.69 + 2% =

226.84 lbs.

6 to 0 = 33.8 ft. x .4575 lbs./ft. + 226.84 lbs. =

242.30 lbs.

TOTAL CHAIN PULL =

242.30 lbs.

When the loads are spaced under 2'-0" on centers, the formula of the total pounds per foot multiplied by the rise in feet can be used to determine the additional force in pounds for the inclined section. This can be seen in the chain pull calculation for the conveyor in Figure #2 from point #3 to #4. It is expressed as (6’-0” x 30 lbs./ft.) or 180 lbs. as the additional force induced by the rise.

In situations where the loads are more than 2'-0" on centers, a more accurate result can be achieved by multiplying the total live load on the inclined portion by the sine of the angle of incline. It will be seen that a maximum of four loads can be on the incline at any one time.

Trigonometry can provide the sine of any angle, but the following angles of slope are most used.

Sine of 15°

=

0.258819

Sine of 30°

=

0.5

Sine of 45°

=

0.707107

Sine of 60°

=

0.866025

Figure #4

14'-0 11/16" Chain @ 3 lbs./ft.

=

42.17 lbs.

4 Pendants & Hooks @ 25 lbs.

=

100 lbs.

4 loads @ 100 lbs.

=

400 lbs.

TOTAL LIVE LOAD

=

542.17 lbs.

542.17 lbs. x 0.707107= 383.37 lbs. Chain Pull

SHORT CUT METHOD OF CHAIN PULL CALCULATION (Approximation Only)

Using a 2.50% friction factor for the short cut method will cover most normal conditions. A large number of vertical and horizontal curves will create slightly higher friction.

CAUTION:

If calculated chain pull using this short cut method is in excess of 550 pounds per drive, you must use the long point-to-point accumulation method of calculation.

STEP #1: DETERMINE QUANTITY OF CARRIERS

The required number of carriers is equal to the total conveyor length divided by the carrier spacing.

STEP #2: DETERMINE QUANTITY OF LOADED AND UNLOADED CARRIERS

Establish distance from loading to unloading. Divide this distance by carrier spacing to determine number of loaded carriers.

STEP #3: DETERMINE LIVE LOAD

The live load on a conveyor is equal to the sum of weights of the chain, pendants, product hook or carrier, and the product.

A. Multiply weight of the chain (3.0 lbs.) by the number of feet of chain.
B. Multiply weight of pendants by required number of total pendants in system.
C. Multiply weight of empty product hooks or carriers by total number of carriers in system.
D. Multiply weight of product only by number of loaded carriers only as determined in Step #2.
E. Total of A, B, C, D = total live load on the conveyor system.

STEP #4 DETERMINE LIFT LOAD

The lift load is the amount of force required to pull the live load upward along the vertical curves in the entire system.

To calculate this force, determine the difference in elevation of all the vertical curves traveling upward in the system. The net vertical rise (in feet) will be considered the total lifting height of the conveyor.

The lift load for the elevation changes of the conveyor is equal to the total lift height (in feet) multiplied by the individual product weight (in pounds), then divided by the load spacing centers in feet.

STEP #5: DETERMINE THE CHAIN PULL

To determine the chain pull due to friction, multiply total moving load by selected friction factor.

A. Multiply total live load from (E) in Step #3 by friction factor of .025.

B. Add lift load obtained in Step #4 to (A) to obtain total chain pull.