Another Catenary Question

[BenAustralia]

Getting myself all confused again!

Typically I find I have a cable I want to know the component reactions at ends of. Lets just assume a hanging cable, equal elevation on supports, gravity load only.

Information I have/can get:

L = Horizontal length of cable
w = weight of cable or UDL on cable, gravity for example
s = Sag (distance at centre from cable to horizontal)

The vertical reactions are fine, purely "w x L / 2". Simple.

The horizontal reaction is something I see conflicting information on.

1. Can I just use "w L ^2 / 2 s" to calculate my horizontal reactions?

2. What would I need to calculate/estimate cable TENSION at ends?

3. What would I need to calculate the Angle at ends? (if i can cal Tension and Angle I can break it back down into components).

Thanks!

 

[CANPRO]

Ben,

I believe your formula for horizontal reactions is correct.

Once you have answered number 1, you have also answered numbers 2 and 3. You have your vertical and horizontal component....just draw your triangle and solve for net tension and angle.

 

[BenAustralia]

Well that part is good to know.

can I derive answers to 2 and 3 using a different formula?

Maybe something to calc the angle at end?

 

[rb1957]

there should be a calc for the slope of a catenary ... worst case wuold be to differentiate the displacement. then since you know the vertical component, you know the resultant.

Quando Omni Flunkus Moritati

 

[paddingtongreen]

The vertical reactions are the same only when the supports are level.
The magnitude and instantaneous slope at the ends can be found using Pythagoras!

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

Actually, the vertical reactions are not w*L/2 for a catenary because the cable is longer than L. If the sag is small, you could make this approximation, but realize that it is an approximation.

If you are using the parabolic approximation, then the horizontal reaction at each end is wL²/8s. That is also the horizontal component in the cable throughout its length.

The tension at the ends (again using the approximation) would be (H+V)^1/2 where H and V are as calculated above. The angle from horizontal is arctan(V/H).

If the cable has appreciable sag, these formulas are not correct. You would have to go to catenary equations which involve hyperbolic functions.

BA

 

[Denial]

I have developed a spreadsheet for the analysis of a cable. It allows for: a single point load somewhere along the cable; the cable ends being at different levels; the cable stretching. It will give you all the information you seem to be looking for. It is available for download from my website

http://rmniall.com

 

[racookpe1978]

Isn't the tension in the cable equal at all lengths of the cable?

 

[Denial]

It is only the horizontal component of the cable tension that is constant along the entire length. (And, obviously, that can only be true if the cable carries no external load that has a non-vertical component.)

 

[BenAustralia]

Thanks BA.

Clears it up.

Say I wanted to use the hyperbolic equations, which ones get me what I want. I see a few with "a" in them, which is just a scaling factor for the shape. How does this relate back to a real world situation?

 

[BenAustralia]

Further to that. Sag is typically about L/12 and looks more parabolic I guess.

Typical smallish tensioned membrane structures.

 

[BAretired]

I am not current with catenary equations. Would have to spend a bit of time studying to answer. Check the link below and Google "catenary curve" for more info.

BA

 

[BenAustralia]

@ Denial.

Not sure on how to use your spreadsheet.

For one, it just gives me an error when running the solver. Runtime error. Macros are enabled.

But the input values I'm not sure of. If I had a cable with no point loads, just simple weight, what would be an example input? Do I need to calculate the "Unstretched Length" of the cable?

 

[BenAustralia]

http://www.meliar.com/design.pdf

Page 21.

For a tension membrane (shade sail) structure, it is saying that Tension in the cable is simply Fabric Prestress x Radius (I can't quite seem to reason this formula in my head though).

Now this obviously makes the assumption that the curve is circular.

So I come to the predicament of if I treat it as Circular, Parabolic or Hyperbolic it seems.

 

[BenAustralia]

I figured it'd make the assumption it was circular, but it does refer to it as Catenary Radius.

 

[paddingtongreen]

It's a second degree parabola. First degree is a beam, second degree takes care of the extra length of cable packed in at the ends. A regular parabola comes very close in a tight wire.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[Denial]

BenAustralia,

The error when running the solver is probably one of two things. Trivial one first: maybe you do not have the solver add-in installed, in which case you will need to fish out your installation disks and install it, then you will need to activate it (by taking Developer>Add-ins, then ticking the appropriate box.)

The other possibility is that you need to "make the solver available" to the VBA environment. This is not the same thing as making the solver available to the Excel environment. Fire up Excel, get into the VBA environment, take Tools>References, find Solver in the list that appears and check its box. (If you cannot find it, you might need to "browse" to it. If your Excel version is <12 you are looking for Solver.xla, if your version is 12 or greater you are looking for Solver.xlam.)

If that doesn't work, see if you can run the Solver from the Excel environment. Enter your problem (or my sample problem). Make sure that your load attachment mechanism is "fixed". Get yourself a solution that is nearly correct (and don't ask me what I mean by "nearly"). Unprotect the worksheet. Fire up Solver. Set your objective cell to Q91, a value to be minimised. Set your changeable variable cells to E15:E18. Do not set any constraints. Run the Solver.

That will at least get you a solution. (To run Solver when the load attachment mechanism is "roller", the cell to be minimised is R91 and the cells that can be changed are E14:E18.)

Now for your other questions. (1) For a cable with no point load, give it a point load of zero somewhere near its midpoint. (2) You DO need to give an unstretched length for your cable, as it this that determines how saggy the cable is.

Please let me know how you go. At the same time tell me which version of Excel you are using, and which version of Windows. I thought I had got around the "VBA reference" problem, but maybe not.

If this issue is going to become a "support" issue, it might save other Eng-Tippers from boredom if you contact me directly, using the e-mail address that is on my web site. but I'm not overly fussed either way.

 

[BAretired]

I figured it'd make the assumption it was circular, but it does refer to it as Catenary Radius.

You are referring to page 21 of the article you referenced earlier. A circular configuration of cable requires that the load be equal along the arc and radial in direction. The sketch would suggest that is not the case. It is not a catenary or parabola either.

Cable configuration must conform precisely to the shape of the bending moment diagram of a simple span beam of length L. Any type of load can be used, uniform, variable, concentrated or any combination of these. Whatever the load, the cable will deflect precisely in accordance with the bending moment diagram.

In particular, a cable which carries a uniform load from end to end will assume the shape of a parabola. The sag may be changed but it will still be a parabola.

A cable which carries nothing but its own weight will assume the shape of a catenary. If the sag is large, the load per foot of span is higher near the supports than at the center.


BA

 

[BenAustralia]

If the cable is carrying a uniform load across its length, much like weight, then it would be a catenary no?

Do you know how the Radius X Stress rule on that page is derived to give the cable tension?

The problem I'm usually presented with is just a tension fabric structure with the fabric almost indicatively drawn in. A beam with uniform UDL gives a parabolic BMD? If so, then I guess stick with the wL^2/2s approximation I've been using?

 

[BAretired]

If the cable is carrying a uniform load across its length, much like weight, then it would be a catenary no?

Yes, if the load is uniform per lineal foot of cable. No, if the load is uniform per lineal foot of span.

Do you know how the Radius X Stress rule on that page is derived to give the cable tension?

Yes. If a cylinder has uniform internal pressure, every part of the circumference feels a radial pressure P. Cutting a section through the middle of the cylinder parallel to the axis in any direction, the total pressure acting on either side of the section is P*D where D is the diameter. Each wall takes half of that, or P*R. It is the same with a cable if it assumes a circular configuration. It is not true if the cable assumes any other configuration.

The problem I'm usually presented with is just a tension fabric structure with the fabric almost indicatively drawn in. A beam with uniform UDL gives a parabolic BMD? If so, then I guess stick with the wL^2/2s approximation I've been using?

If the load is uniform, the cable is parabolic and the horizontal component of the tension is wL²/8s...not wL²/2s as you just stated. Whether or not the cable is parabolic, the horizontal component of the cable tension is M/s where M is the simple span moment at any point and s is the sag at the same point.

The maximum cable tension occurs at the supports and as should have been stated earlier is (V²+H²)^1/2. Please note I had a typo in my earlier post.

BA