Cable Loads

Force and tension in cables supporting uniformly loads

cable load force tension

The midspan force in a cable, wire or rope supporting a uniformly distributed load can be expressed as

H = w L2 / (8 d)      (1)

where

H = midspan force in the cable (lb, N)

w = unit load (weight) on the cable (lb/ft, N/m)

L = cable span (ft, m)

d = cable sag (ft, m)

The forces at the support ends of the cable can be estimated as

T = (H2 + (w L / 2)2)0.5    (2)

where

T = forces at supports (lb, N)

The length of cable can be approximated to

S = L + 8 d2 / (3 L)       (3)    

where

S = cable length (ft, m)

 

  • kip = 1000 lb
  • klf = kip per linear foot

Example - Uniform Cable Load, Imperial units

A cable with length 100 ft and a sag 30 ft has a uniform load 850 lb/ft. The midspan force in the cable can be calculated as

H = (850 lb/ft) (100 ft)2 / (8 (30 ft))

   = 35417 lb 

The force at the supports can be calculated as

T = ((35417 lb)2 + ((850 lb/ft(100 ft) / 2)2)0.5  

  = 55323 lb

The length of the cable can be estimated to

S = (100 ft) + 8 (30 ft)2 / (3 (100 ft))

   = 124 ft

Example - Uniform Cable Load, SI units

A cable with length 30 m and a sag 10 m has a uniform load 4 kN/m. The midspan force in the cable can be calculated as

H = (4000 N/m) (30 m)2 / (8 (10 m))

   = 45000 N

   = 45 kN 

The force at the supports can be calculated as

T = ((45000 N)2 + ((4000 N/m(30 m) / 2)2)0.5  

  = 75000 N

  = 75 kN

The length of the cable can be estimated to

S = (30 m) + 8 (10 m)2 / (3 (30 m))

   = 38.9 m

Related Topics

  • Mechanics - Forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more
  • Statics - Loads - force and torque, beams and columns

Related Documents

Tag Search

  • en: cable loads tension uniform wire rope
  • es: cable de tensión de la cuerda de alambre cargas uniforme
  • de: Kabellasten Spannung gleichmäßige Seil

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