Tension in a Wire: Calculating Maximum Load Capacity

[bobbles22]

Hi Guys,

Firstly, I'm not a school student, I work for a cable company. This is just a question I was wanting some verification on regarding a situation for a quote we have in.

We have a client wanting 50m of a cable to hang clothes on in a warehouse. The breaking load of the cable (if hung from a single point vertically) is 5670kg (56700N). If it is hung horizontally (tensioned to a point where the cable is effectivly horizontal, not much more), what is the maximum load it can take? He only wants to hang about 200kg of clothing on it. My head says it should be absolutly fine. I've got a Young's Modulus for the cable of 6500 kg/mm². The cross-section of the cable is about 46.5mm².

Any thoughts? It seems to my common sense head that it should easily be able to hold it. I'm just wondering what it would be able to hold.

Thank you

 

[Matterwave]

So, from a pure physics point of view, the tension can be calculated easily as a function of the angle theta that you want the cable to make with the horizontal at the two ends.

Most simply, one neglects the mass of the cable itself, drawing a force diagram, you will find that the tension is required to balance the load in the center: 2Tsin(theta)=mg so that T=mg/(2sin(theta)), noticing that the cable cannot be perfectly horizontal. You can be conservative and put the mass of the cable in the "m" term in that equation.

For example, if your load is 200kg, and your cable is say 30kg, and the angle theta is say 1 degree, then T~1400N (not even close to your 56700 figure), and for a theta of .1 degree, T~11,500N (still far from that figure).

Now, this analysis is from a purely physics point of view (as opposed to an engineering POV), I'm not sure how the sheer stress in this system would be handled by the cable.

 

[jambaugh]

The devil is in the details of "effectively horizontal". To get it perfectly horizontal requires infinite tension.

For a ball-park guesstimate figure all the cloth bunched in the middle. You can work out the maximum load for a given sag as a percent of span or the minimum sag for a given load.

If you hang a weight in the middle of a wire, the tension will be:
$$T = \frac{wL}{4d}$$
where T is tension, w is weight hung, L is the length of wire and d is the vertical droop or drop. You have the weight held by both sides so you have equal ratios of w/2:T = d:L/2

This discounts the weight of the wire.

So if say you're hanging 200kg with a tension of 5000kg weight you'll have a % drop of d/L = w/4T = 1/10000. That's 1cm per 100 meters.

I would think that is amply straight for your client's purposes. And of course distributing the weight will reduce the amount droop at the middle, improving things a bit.