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Wire Rope Slings

Reference data and engineering information about wire rope slings for electrical applications.

wireropeslings

Overview

Engineering reference data for Wire Rope Slings in electrical engineering.

Key Formulas

Ohm's Law

V=IRV = IR

Voltage = Current × Resistance.

Power

P=VI=I2R=V2/RP = VI = I^2R = V^2/R

Electrical power.

Energy

E=PtE = Pt

Energy = Power × Time.

Variables

Symbol Description Unit
VV Voltage V
II Current A
RR Resistance Ω
PP Power W

Reduction Factors for Sling Angles

Sling Configurations

Wire rope slings can be arranged in multiple configurations depending on the lifting application:

  • Two Slings: Most common configuration; requires input of load mass, sling spread length (a), and lifting height (h)
  • Three Slings: Used for heavier or irregularly shaped loads; provides better load distribution
  • Four Slings: Used for large, heavy loads; requires two spread dimensions (a and b) plus lifting height (h)

Practical Examples

Single Vertical Sling

For a body with mass m=100m = 100 kg suspended by a single vertical sling:

F=W=mg=(100 kg)(9.81 m/s2)=981 N=9.8 kNF = W = mg = (100 \text{ kg})(9.81 \text{ m/s}^2) = 981 \text{ N} = 9.8 \text{ kN}

Two Vertical Slings

For a body with weight W=1000W = 1000 N supported by two vertical slings:

F=W2=1000 N2=500 N=0.5 kNF = \frac{W}{2} = \frac{1000 \text{ N}}{2} = 500 \text{ N} = 0.5 \text{ kN}

Two Slings at 30° from Vertical

For a body with weight W=1000W = 1000 N supported by two slings at α=30°\alpha = 30° from vertical:

F=W/2cos⁡(30°)=W/2f=1000 N/20.866=577 N=0.58 kNF = \frac{W/2}{\cos(30°)} = \frac{W/2}{f} = \frac{1000 \text{ N}/2}{0.866} = 577 \text{ N} = 0.58 \text{ kN}

The effective capacity is reduced to 86.6%86.6\% of the vertical configuration at this angle.

Interactive Charts

References