• posted

I need (well, want) to compute the height of loads (specifically, traffic signals and signs) hung from a suspension cable. Googling found a lot of catenary stuff (of which I was already somewhat aware), and some about uniform loading, but (AFAICT) nothing helpful.

I found a book with a solution of sorts (Peterson, 'Applied Engineering Mechanics'). It treats the cable as straight line segments between the load attachment points. I don't yet know whether that's an acceptable model. Even if it is, Mr. Peterson seems to insist that the coordinates (x,y) of one of the (two) end-most loads be known in advance. This is not out of the question, but it would be easier if either the total cable length, or the coordinates of the 'lowest' load, could be provided instead.

I'd appreciate any clues.

TIA, George

• posted

I have a text, "Time-Saver Standards for Building Types", DeChiara and Callender, McGraw-Hill, ISBN

0-07-016218-2

On page 703 the subject is Truck Dealer and Service Facilities. In the article, they list the service doors as being 14 feet high. However, if I remember correctly, because trucks do carry over sized loads, many under passes are 16 feet high. You can check with the Department of Transportation in your state to verify this.

As far as the cable sag, I suggest you get information from someone who sells cable. Years ago I had a catalog from Bethlehem Steel cable division in which they included example problems and the recommended methods of solution. I remember that steel cable does stretch in tension and that catalog listed the percentage based on the load.

If I can locate that catalog (I doubt it though), I will get back to you.

Jim Y

• posted

Callender, McGraw-Hill,

article, they list the

many under passes are

state to verify this.

cable. Years ago I

example problems and the

tension and that

Jim Y

• posted

Jim Y

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Well, maybe parts of an answer. From sci.math, I got a reference (Irvine, 'Cable Structures'), which I'm getting via inter-library loan. It comes highly recommended. Whether it's approachable is yet to be seen.

Also from sci.math, I got the suggesting that "Minimizing height of center of gravity under the constraint of fixed cable length and assuming cable weight is zero can give you an approximate solution," and a Mathmatica program to do that. Unfortunately, I don't have Mathmatica, and it has some operators/functions that (a) aren't intuitively obvious to me and (b) look like they'd be a considerable exercise to translate into C (or whatever).

My current thinking is to treat it as a zero-weight cable. (Field observations seem to be that the span wire segments are straight lines.) The easiest contraint probably is a fixed cable length. With that, and an some initial estimates, I should be able to iterate until I get a configuration where the forces and moments are in balance. There remain a few details to fill in.

Or something like that.

George

• posted

Let me try to get you started. You have two posts that are separated by a distance of "L". The weight "W" is located in the exact center of "L". Your cable has a Tension of "TL" in the left and "TR" in the right and are the hypotenuse of *2* right triangles. ("TL" = "TR") The left right triangle consists of a vertical force of W/2 and a hypotenuse force (tension) of "TL". From those you can get the horizontal force. Now you have a right triangle of three forces that equate to ... you can take from here. Two sides of a right triangle give you the third and from there you can get the angles. Draw the picture. The forces must be in equilibrium. Remember you must set the tension based on the cable as I stated earlier. It is called an *assumption* for theoretical purposes, but a cable catalog can give you a real number. This method may be "trial and error", but it will help you to select the cable, determine the height that the cable is fastened to the support posts and know how far the cable sags due to the weight. Sin = a/c Cos = b/c Tan = a/b = Sin/Cos Sin = (W/2)/(TL)

May I suggest you look at a copy of "3000 Solved Problems in Physics", Alvin Halpern, Schaum's Solved Problem Series, ISBN 0-07-025636-5, Chapter 2, Problem 2.7. This problem has unequal legs for "L", but you should be able to solve your problem using it as a guide.

Jim Y

• posted

distance of "L". The

"TL" in the left

"TR") The left right

of "TL". From

three forces that equate

third and from there

Remember you must

*assumption* for

method may be "trial

that the cable is

weight. Sin = a/c Cos

Halpern, Schaum's

problem has unequal legs

• posted

This is a solveable problem, although it will need to be solved for each case (the specific number of loads, spacings, etc.) that you want to consider. It is a bit of a bear, because it involves fitting the catenary equation between the load points and summing forces to zero at each load point. This is not impossible, just tedious. How bad do you want the answer? -- Sam

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