Cable Loads

Force and tension in cables with uniform loads

Uniformly Loaded Cables with Horizontal Loads

Uniformly loaded cable with horizontal chords

The midspan cable and horizontal support forces with uniformly distributed load can be calculated as

R1x = R2x = Rx = q L2 / (8 h)                               (1)

where

R1x = R2x = Rx midspan cable or horizontal support forces  (lb, N)

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

L = cable span (ft, m)

h = cable sag (ft, m)

The vertical support forces at the end of the cable can be calculated as

R1y = R2y = Ry = q L / 2                               (2)

where

R1y = R2y = Ry = vertical support forces  (lb, N)

The resultant forces acting in the support ends - and in the direction of the cable close to the supports -  can be calculated as

Tmax = (Rx2 + Ry2)0.5                               (3)

where

Tmax = resultant force at the support (lb, N)

The length of cable can be approximated to

S = L + 8 h2 / (3 L)                                 (4)    

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 horizontal supports and midspan cable forces can be calculated as

R1x = R2x = (850 lb/ft) (100 ft)2 / (8 (30 ft))

        = 35417 lb 

The vertical forces at the supports can be calculated as

R1y = R2y = (850 lb/ft) (100 ft) / 2

        =  42500 lb

The resultant forcee acting in the supports can be calculated as

Tmax = ((35417 lb)2 + (42500 lb)2)0.5    

   = 55323 lb

The length of the cable can be approximated 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 of 4 kN/m. The horizontal supports and midspan cable forces can be calculated as

R1x = R2x = (4000 N/m) (30 m)2 / (8 (10 m))

   = 45000 N

   = 45 kN 

The vertical support forces can be calculated as

R1y = R2y = (4000 N/m) (30 m) / 2

              =  60000 N

              = 60 kN

The resultant forcee acting in the supports can be calculated as

R = ((45000 N)2 + (60000 N)2)0.5    

           = 75000 N

           = 75 kN

The length of the cable can be approximated to

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

   = 38.9 m

Uniformly Loaded Cables with Inclined Chords

Uniformly loaded cable with inclined chords

The horizontal cable and support forces with uniformly distributed load can be calculated as

R1x = R2x = Rx = q L2 / (8 h)                               (5)

The vertical support forces at the end of the cable can be calculated as

R1y = R2y = Ry = Rx d / L + q L / 2                               (6)

The resultant forces acting in the support ends - and in the direction of the cable close to the supports -  can be calculated as

Tmax = (Rx2 + Ry2)0.5                               (7)

Related Topics

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

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