Hoisting and Cable Tensioning with Cylinders

Figure 4 – Suspended Load on a Cable.

This is a simplified method of finding cylinder force for tensioning a cable, and was derived from trigonometry, although a knowledge of trig is not necessary in order to use the method. Ratio values in the left column of the chart are “cotangent” values taken from a standard table of natural trigonometric functions. Multipliers in the right column are “cosecant” values for the same angle. For intermediate ratio values or those beyond the chart, regular trig tables may be used.

Assume the weight of the cable is relatively small as compared with the suspended weight. Tension in each leg of the cable will be equal if the weight is hanging in the exact center of the cable. Otherwise, the tension will be different in each leg. Solve for each leg.

The first step, always, is to determine what percent of the load is supported by each leg of the cable. The load on each leg will be inversely proportional to distance A or A’ from weight to support. In Fig. 4, total span is 20 feet, so 9,000 lbs. (75% of the weight) will be supported by the short leg and 3,000 lbs. (25% of the weight) will be supported by the longer leg.

To solve for tension in a leg, after finding what per cent of the load it carries, take distance-A and divide it by the distance B (the sag) to find the ratio A ÷ B. Enter the left column of the chart and find this ratio. Use the multiplier from the right column to multiply times the portion of load carried by that leg. This gives the cylinder force necessary to support the load in this position.

If the exact value of your calculated ratio does not appear in the chart, interpolate between the next higher and next lower ratio; then interpolate in the same way to find the multiplier.

Solve for tension in short leg (Fig. 4): Ratio A ÷ B = 5 ÷ 2 = 2.50. Enter left column of chart and get multiplier, 2.69, on same line as ratio 2.50. Tension = 2.69 x 9,000 lbs. = 24,210 lbs.

Solve for tension in longer leg: Ratio A ÷ B = 15 ÷ 2 = 7.50. Enter left column of chart and find ratio 7.50. Use multiplier 7.57 shown on same line. Tension= 7.57 x 3,000 lbs. = 22,710 lbs.

In sizing a cylinder to tension the above cable, do not add tension in both legs. One cylinder will tension both legs of the cable. If pulling on the short leg, it must have 24,210 lbs. force; if pulling on the longer leg it must have 22,710 lbs. force.

A small difference in elevation between the two end supports will not seriously affect the accuracy of tension calculations. However, if the difference in elevation results in an angle between cable and horizontal of more than 10 to 15°, a more involved trigonometric calculation will have to be made.

Hydraulic Motor Torque…

When pulling tension on a loaded cable with a winch drum driven with a hydraulic motor, use procedure described above for finding cable tension. Then solve for torque on winch drum by taking tension times drum effective radius. Then calculate theoretical motor torque by dividing drum torque by ratio between drum sheave or sprocket and hydraulic motor sheave or sprocket.

Example: Cable tension = 15,000 lbs; winch drum diameter 14 inches; winch drum drive sprocket diameter= 2 inches; motor sprocket diameter = 6 inches. Find theoretical torque required from the hydraulic motor.

Solution: Winch drum torque is 15,000 x 7 (radius of drum) – 105,000 inch-lbs. Motor torque is found by dividing winch torque by drive sprocket ratio. Torque= 105,000 / 2 = 52,500 inch-lbs. This is theoretical torque required on the motor shaft. Allow extra torque for mechanical losses in sprocket drive and in winch drum bearings.