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Cable Loads

Force and tension in cables with uniform loads

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Uniformly Loaded Cables with Horizontal Loads

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

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)

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