Need Formula to Calculate Stress on a Suspended Line

Thanks to everyone so far. Sparkstation, you have put me on the right track, but I might need to do some more digging around because I'm not sure that your reference actually answers my question. First, I am not using any "hangers" and second, I would have thought T=tension would vary depending upon the point along the cable yet there doesn't appear to be any place in the formula to input that point -- but maybe I'm just misunderstanding the whole concept. To illustrate my point, imagine a long chain with an extreme amount of sag in it relative to span, to the point that the chain is _almost_ folded in two and hanging straight down; the tension at each end would approach half of the weight of the chain, the tension in the middle virutally nothing, and tension anywhere in between as you move up or down the chain. On the other hand, a chain stretched _completely_ taut ... (not actually possible, but 'theoretical' for my point) ... across a span with no sag whatsoever would have tension that would be equal all along the chain and would be many times the weight of the chain. Or am I completely wrong about all of that? Anyway, I'm somewhere in between, but all I really need to know is the greatest amount of tension, which I believe should be at the ends of the span. Does the formula provided in your reference calculate the tension at the ends?

Horrible Old Bat, is there a way for me to plug that function into something to calculate tension?

Ozzb, my vine is a hop vine that is cut down at the end of the season for harvesting hop cones, so the twine need last only one season; cable or wire is cost prohibitive in that application. Otherwise, you are absolutely correct, and I appreciate your input.

Last year my first-year plants only partially spanned the arbor, and did not have a lot of weight on them at the top of the plants (the portions beginning to grow across the span). This year should produce more growth (they start out at ground level on new twine), and next year they should span the entire arbor (hops grow as much as 25 feet in one year). The vines themselves have a lot of strength that they will bring with them as they grow, but until they completely cross the span and firmly anchor themselves on the other side of the arbor, all of the tension created by their weight must be borne soley by the twine on the far side. I use hay-baling twine which has been sufficient to support nine feet of growth on a 60 degree angle from the ground. I had a 12' span between trellises last year, but am thinking of reducing that to 10'. The trellises are 8' tall, so I'd prefer not to have more than about 1' of sag in the middle so there will still be clearance to walk under them.

By the way, is this forum considered to be the most appropriate one for my questions?

Thanks for the help, and for any additional info any one can provide.

Bill Velek

It forms the shape described by the function y = A cosh(Bx) + C.

Twine and rope are not a good choice for long term exposer to the elements. the degrade over time that will cause your vines to fall away. I would recommend you look at aircraft cable(wire rope). In some of the small diameters they have them with colored PVC coverings. The green works well to blend in. You will get may years of reliable service and will not have to worry about stress on the line.

The catenary is theoretically correct if the load is uniformly distributed along the length of cable. If you are spanning ten or twelve feet with only a foot of sag, then you are close enough to assume a parabolic drape. In fact, you would be precisely correct if your load were uniformly distributed along the span rather than the cable length.

The bending moment, 'M' for a span of 'L' is wL²/8. The tension in the cable is then M/s where s = sag.

Thus, the tension, near enough for all practical purposes is T = wL²/8s.

For a span of 10', a sag of 1' and a load of 20 #/', the tension would be 250 pounds.

The tension is uniform over the full length of the cable using this assumption.

"The tension is uniform over the full length of the cable using this assumption."

Sorry, that is not quite correct. What I meant to say was that the horizontal component of the tension is uniform over the full length of cable. The tension in the cable is maximum at the point of greatest slope, i.e. at the supports. In the case cited above, the tension at the supports is about 8% greater than the tension in the middle.

The end reactions each have a vertical component of 5w or 100#. The horizontal component at each end is 250#, so the cable tension is 269# at each end.

Probably draw a line touching top ends of the trellises and parallel to their bases. now from top of each trellise draw another line to centre of their bases. this way you may get a triangle that makes force distribution along its each arm. then apply trigonometry like cos theta etc for force distribution along twine/rope side. perhaps you may be able to solve your problem with this trick.

man singh

"Roark's Formulas for Stress and Strain" published by McGraw-Hill will have the answer you are looking for. It should be available at most libraries and could be considered to be a stress analysist's bible.