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## Force and tension in cables with uniform 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)

## Related Topics

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

## 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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## Citation

• Engineering ToolBox, (2012). Cable Loads. [online] Available at: https://www.engineeringtoolbox.com/cable-loads-d_1816.html [Accessed Day Mo. Year].

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