# Cable Loads

## Force and tension in cables supporting uniformly loads

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The midspan force in a cable, wire or rope supporting a uniformly distributed load can be expressed as

*H = w L ^{2} / (8 d) (1)*

*where *

*H = midspan force in the cable (lb, N)*

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

*L = cable span (ft, m)*

*d = cable sag (ft, m)*

The forces at the support ends of the cable can be estimated as

*T = (H ^{2} + (w L / 2)^{2})^{0.5} (2)*

*where *

*T = forces at supports (lb, N)*

The length of cable can be approximated to

*S = L + 8 d ^{2} / (3 L) (3) *

*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 midspan force in the cable can be calculated as

*H = ( 850 lb/ft) (100 ft)^{2} / (8 (30 ft))*

* = 35417 lb *

The force at the supports can be calculated as

*T = (( 35417 lb)^{2} + ((850 lb/ft) (100 ft) / 2)^{2})^{0.5} *

* = 55323 lb*

The length of the cable can be estimated 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 *4 kN/m*. The midspan force in the cable can be calculated as

*H = (4000 N/m) (30 m) ^{2} / (8 (10 m))*

* = 45000 N*

* = 45 kN *

The force at the supports can be calculated as

*T = (( 45000 N)^{2} + ((4000 N/m) (30 m) / 2)^{2})^{0.5} *

* = 75000 N*

* = 75 kN*

The length of the cable can be estimated to

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

* = 38.9 m*

## Related Topics

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