# Cable Loads

## Force and tension in cables with uniform loads

The equations below can also be used for cables only loaded with their own weight as long as the sagging height *(h)* vs. length *(L)* ratio is below *0.1*.

### Uniformly Loaded Cables with Horizontal Loads

The cable follows the shape of a parable and the horizontal support forces can be calculated as

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

* = q L ^{2} / (8 h) (1)*

*where *

*R _{1x} = R_{2x} = *horizontal support forces (lb, N) (equal to midspan lowest point tension in cable)

*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} *

* = q L / 2 (1a)*

*where *

*R _{1y} = R_{2y} = *vertical support forces (lb, N)

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

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

* = (R _{1x}^{2} + R_{1y}^{2})^{0.5} *

* = (R_{2x}^{2} + R_{2y}^{2})^{0.5} (1b)*

*where *

*R _{1,2}* = resultant force at the support (lb, N)

The angle θ can be calculated as

*θ = tan ^{-1}(R_{1y }/ R_{1x}) *

* = tan ^{-1}(R_{2y }/ R_{2x}) (1c)*

The length of the sagged cable can be approximated to

*s = L + 8 h ^{2} / (3 L) (1d) *

*where *

*s = cable length (ft, m) *

Note that the equation is not valid when h > L / 4.

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

#### Uniformly Loaded Cables with Horizontal Loads - Calculator

*q - uniform load (N/m, lb/ft)*

* L - length (m, ft)*

* h - sag (m, ft)*

#### 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 forces acting in the supports can be calculated as

*R _{1,2}* = ((

*35417 lb*)

^{2}+ (

*42500 lb)*

^{2})

^{0.5}

* = 55323 lb*

The angle *θ* can be calculated as

*θ = tan ^{-1}((42500 lb) / (35417 lb)) *

* = 50.2 ^{o} *

The length of the sagged 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 angle *θ* can be calculated as

*θ = tan ^{-1}((60 kN) / (45 kN)) *

* = 53.1 ^{o} *

The resultant force acting in the supports can be calculated as

*R _{1,2} = ((45000 N)^{2} + (60000 N)^{2})^{0.5} *

* = 75000 N*

* = 75 kN*

The length of the sagged cable can be approximated to

*s = (30 m) + 8 (10 m) ^{2} / (3 (30 m))*

* = 38.9 m*

#### Example - Known Tension at the Supports - Calculate Sagging and Length of Cable

For a *30 m* long cable with uniform load *4 kN/m* the resulting tension in the cable at the end supports are *100 kN*.

The vertical forces in the supports can be calculated as

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

* = **( 4 kN/m)*

*(30 m)*/ 2

* = 60 kN*

The horizontal forces in the supports can be calculated as

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

* =* ((100 kN)

^{2}- (60 kN)

^{2})

^{0.5}

* = 80 kN *

The angle *θ* can be calculated as

*θ = tan ^{-1}((60 kN) / (80 kN)) *

* = 36.9 ^{o} *

The sagging can be calculated by modifying eq.1 to

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

* = (4 kN/m) (30 m) ^{2} / (8 (80 kN))*

* = 5.6 m *

The length of the sagged cable can be estimated to

*s = (30 m) + 8 (5.6 m) ^{2} / (3 (30 m))*

* = 32.8 m*

* *

### Uniformly Loaded Cables with Inclined Chords

If heights *h _{1}* and

*h*are known the horizontal support forces can be calculated as

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

* = q L ^{2} / (2 ((h_{1})^{0.5} + (h_{2})^{0.5})) (2)*

If distance *a and b* are known - the horizontal support forces can be expressed as

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

* = q a ^{2} / (2 h_{1}) *

* = q b^{2} / (2 h_{2}) (2b)*

If *b > a* the maximum forces in the cable and at support *1 and 2* can be calculated as

*R _{2} = (R_{2x}^{2} + (q b)^{2})^{0.5} (2c)*

*R _{1} = (R_{1x}^{2} + (q a)^{2})^{0.5} (2d)*

- and the vertical forces at support *1 and 2 *can be calculated as

*R _{2y} = (R_{2}^{2} - R_{2x}^{2})^{0.5} (2e)*

*R _{1y} = (R_{1}^{2} - R_{1x}^{2})^{0.5} (2f)*

The angles between the horizontal and resulting forces can be calculated as

*θ _{2} = cos^{-1}(R_{2x }/ R_{2}) (2g)*

*θ _{1} = cos^{-1}(R_{1x }/ R_{1}) (2g)*

The length of the sagged cable can be estimated as

*s _{b} = b (1 + 2/3 (h_{2 }/ b)^{2}) (2h)*

*s _{a} = a (1 + 2/3 (h_{1 }/ a)^{2}) (2i)*

*s = s _{a} + s_{b} (2j)*

#### Example - Inclined Cable with Uniform Load, SI units

A cable with length *30 m* and a sag *h _{2} = 10 m* and

*h*has a uniform load of

_{2}= 1 m*4 kN/m*.

The horizontal support forces can be calculated as

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

* = (4 kN/m) (30 m) ^{2} / (2 (((1 m))^{0.5} + ((10 m))^{0.5})) *

* = 103.9 kN *

The distance *a* and *b* can be calculated by rearanging eq. *2b* to

*a = (2 R _{1x} h_{1} / q)^{0.5}*

* = (2 (103.9 kN) (1 m) / (4 kN/m)) ^{0.5} *

* = 7.2 m*

*b = (2 R _{2x} h_{2} / q)^{0.5}*

* = (2 (103.9 kN) (10 m) / (4 kN/m)) ^{0.5} *

* = 22.8 m*

The resulting forces in the supports can be calculated as

*R _{2} = ((103.9 kN)^{2} + ((4 kN/m) (22.8 m))^{2})^{0.5} *

* = 138.2 kN *

*R _{1} = ((103.9 kN)^{2} + ((4 kN/m) (7.2 m))^{2})^{0.5} *

* = 107.8 kN*

The vertical forces in the supports can be calculated as

*R _{2y} = ((138.2 kN)^{2} - (103.9 kN)^{2})^{0.5} *

* = 91.2 kN *

*R _{1y} = ((107.8 kN)^{2} - (103.9 kN)^{2})^{0.5} *

* = 28.8 kN *

The angles between resulting and horizontal forces in support 1 and 2 can be calculated as

*θ _{2} = cos^{-1}((103.9 kN)/ (138.2 kN)) *

* = 41.3 ^{o} *

*θ _{1} = cos^{-1}((103.9 kN)/ (107.8 kN)) *

* = 15.5 ^{o} *

The length of the sagged cable can be calculated as

*s _{b} = (22.8 m) (1 + 2/3 ((10 m) / (22.8 m))^{2}) *

* = 25.7 m *

*s _{a} = (7.2 m) (1 + 2/3 ((1 m) / (7.2 m))^{2}) *

* = 7.3 m*

*s = (7.3 m) + (25.7 m ) *

* = 33 m *

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