How much force (in lbs) is required to stretch a wire between poles. (sag = 3')?

Question

The wire is 150' long, and weighs 75 lbs. I'm looking for the force (in pounds) to stretch the wire, allowing a 3 foot deep sag in between the poles used to hang the wire on.

The Wire is 150' long, and weighs 75 lbs. Ambient temp: 90 deg F. There is no ice or other loads such as wind on the wire for the purpose of this question.

The exact issue is how much force is required to lift a wire that is fastened to one pole at say 10' off the ground, and stretch it to toward a second poles tie point, at the same 10' off the ground, allowing a final sag of 3' on the wire. Again, the total weight of the wire is 75lbs, but the wire is 150' long.

Thanks!

Answer 1

I found a sag table for 477 aluminum conductor (code named COSMOS). 150' of this conductor weighs 67 lbs. (pretty close to your 75 lbs)

At 90 F, 477 AL wire will have an initial sag of about 2' and a tension around 650 lbs. Allowing it to sag 3' will have substantially lower tension. Over time the wire will stretch a little bit. The final sag will be closer to 31" and have a tension closer to 500 lbs.

Keep in mind that the poles will need to support this tension without leaning. As the poles lean towards one another, the sag increases dramatically.

650 lbs x 10 ft = 6500 lb-ft of bending moment at the ground line.

Answer 2

It depends on the unit weight of the wire
P = w L^2 / 8 a = W L / 8a . . . where w = WL
P = 75 ( 150 ) / (8 x 3) = 468.75 lbs