Selection of steel wire rope in hoisting operation

The steel wire rope is the most important part of the common lifting equipment and maintenance, and it is very important to ensure the correct connection between the steel wire rope and the lifting equipment.

1、 Performance introduction of steel wire rope

Steel wire rope is generally made of high carbon steel wire with many specifications, but generally, our commonly used specifications are: 6 × 19+FC（IWR）、6 × 37+FC（IWR）、6 × 61 + FC (IWR), three specifications. Among them, 6 represents the number of steel wire strands, 19, 37 and 61 represent the number of steel wires per share, "+" represents the rope core in the middle of 6 strands, IWR represents the steel core and FC represents the fiber core.

Under the condition of the same diameter

Wire rope 6 × 19 + FC (IWR): the steel wire has large diameter, high strength, but poor flexibility. It is often used as cable wind rope.

Wire rope 6 × 37 + FC (IWR): the diameter of steel wire is general, the strength is general, but the flexibility is general. It is often used as sling.

Wire rope 6 × 61 + FC (IWR): the steel wire is the thinnest, with low strength but good flexibility. It is often used as a sling.

2、 Selection method of steel wire rope:

1. A concept, safety factor (k), must be introduced here. The safety factor of steel wire rope refers to the reserve tensile force that is allowed to bear the tensile force in the specified use of steel wire rope, that is, the rated margin when the steel wire rope is broken in use.

2. Generally speaking, as a sling, the safety factor (k) of steel wire rope is "8-10", and for carrying people, the safety factor is "12-14".

3. Steel wire rope is selected, and the calculation of allowable tension T:

The allowable tension must be greater than the weight of the lifted object, and the calculation formula is:

T=P/K

Allowable tension = breaking force of steel wire rope (kg, N, MPA) / safety factor

The breaking force of steel wire rope (kg, kn, MPA) can be obtained by referring to the performance table provided by the steel wire rope manufacturer.

A kind of steel wire rope commonly used in our construction site 6 × 37 + 1 (IWR), nominal tensile strength 170kg / mm2.

For example: a 10 ton equipment needs to be hoisted on site, with double strands under stress. What diameter of steel wire rope should be selected?

The calculation process is as follows:

10 ton equipment, double strand stress, single strand stress of 5 tons, safety factor of 8, then

Breaking force = 5000kg × 8=40000kg，

Sectional area of steel wire rope = breaking force ÷ nominal tensile strength = 40000 / 170 = 235.29mm2.

At this time, according to the above table, the selected cross-sectional area is 250.95 > 235.29, and the corresponding wire rope diameter is 26mm

Note that the diameter of the steel wire rope cannot be deduced from the cross-sectional area, because there is a gap between the steel wires, which is bound to cause the selected diameter to be too small.

For example, as the management party, we check how to calculate the steel wire rope.

Is it feasible for the construction unit to use steel wire rope with diameter of 19.5mm, four strands of force and hoist equipment weighing 10 tons?

According to the table or calculation, the breaking force of single strand steel wire rope is p = 23950kg, and the safety factor K is "8".

Then the allowable tension of single strand steel wire rope is t = 23950 / 8 = 2993.75kg. According to the above calculation, the allowable tension of single strand is 2993.75 and that of four strands is 2993.75 × 4 = 11975kg, meeting the requirements.

Summary:

1. As a sling, the safety factor (k) of steel wire rope is "8-10" (this is remembered as common sense).

2. Allowable tension = breaking force of steel wire rope (kg, N, MPA) / safety factor

3. Field experience formula:

Breaking force of steel wire rope (kg) = D ²× 52 (there is no table to check on site, and the empirical method can be used)

Example: the diameter of steel wire rope is 19.5mm, and the breaking force is 19.5 × nineteen point five × 52=19773kg。