General question to all the Stress guys: I wounder if this is applicable......:

Cable Loads

Tension in cable supporting uniformly load

cable load

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

H = w L2 / (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)

mass and weight

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

T = (H2 + (w L / 2)2)0.5 (2)

where

T = force at supports (lb, N)

The length of cable can be approximated to

S = L + 8 d2 / (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

http://www.engineeringtoolbox.com/cable-loads-d_1816.html