# 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

*R _{1x} = R_{2x} = R_{x} = q L^{2} / (8 h) (1)*

*where *

*R _{1x} = R_{2x} = R_{x} = * 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

*R _{1y} = R_{2y} = R_{y} = q L / 2 (2)*

*where *

*R _{1y} = R_{2y} = R_{y} =* 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

*T _{max} = (R_{x}^{2} + R_{y}^{2})^{0.5} (3)*

*where *

*T _{max}* = resultant force at the support (lb, N)

The length of cable can be approximated to

*S = L + 8 h ^{2} / (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

*R _{1x} = R_{2x} =* (

*850 lb/ft*) (100 ft)

^{2}/ (8 (30 ft))

* = 35417 lb *

The vertical forces at the supports can be calculated as

*R _{1y} = R_{2y} = *

*(*

*850 lb/ft*)*(100 ft)*/ 2

* = 42500 lb*

The resultant forcee acting in the supports can be calculated as

*T _{max}* = ((

*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

* R_{1x} = R_{2x} =* (4000 N/m) (30 m)

^{2}/ (8 (10 m))

* = 45000 N*

* = 45 kN *

The vertical support forces can be calculated as

*R _{1y} = R_{2y} = *

*(*

*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

*R _{1x} = R_{2x} = R_{x} = q L^{2} / (8 h) (5)*

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

*R _{1y} = R_{2y} = R_{y} = R_{x} 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

*T _{max} = (R_{x}^{2} + R_{y}^{2})^{0.5} (7)*

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