Hanging a point load from a catenary

[Medgy]

Homework Statement

I have attached a scanned file of the problem outline, including all known data. For the initial curve, H, (horizontal force) is constant, while for the final curve, H', is a combination of the initial H and the stress added by the point load, P.

I run into issues trying to solve this problem because the addition of the point load changes Xc and Yc in ways which I can't calculate (and therefore can't define the final catenary). A good start would be calculating H', but I am also unclear how to do this since I don't know the angles at the supports.

Homework Equations

Included in image.

The Attempt at a Solution

I would even be happy with a simpler solution at first where the point load is applied at the initial Xc and Yc, allowing the catenary to be split into two catenaries, but I don't know how the load affects the sag, or Yp, which means I don't know the co-ordinate required to split up the two curves.Really, any help is appreciated! Thanks.

 

[phinds]

I have not really looked at your post beyond seeing that you have a load hanging from what started out as a catenary and for some reason you think that the resultant curve will ALSO be a catenary. I don't think so. A catenary is a curve that results from a uniformly distributed length of wire/chain/rope/whatever.

My understanding is that suspension bridges start out with a heavy hanging wire bundle that is a catenary and they go to great lengths to KEEP it in a catenary shape by carefully distributing the load along the vertical wire bundles.

A single load at a single point will distort the curve and I don't know what you end up with but it is not a catenary.

 

[Medgy]

The curve from the left support to the load will be a catenary, and the curve from the load to the right support will also be a catenary. The entire curve will need to be split in two and defined separately. The major obstacles are calculating the new horizontal tension in the wire for each section, and calculating the point in the y co-ordinate that the mass hangs.

 

[phinds]

The curve from the left support to the load will be a catenary, and the curve from the load to the right support will also be a catenary. The entire curve will need to be split in two and defined separately. The major obstacles are calculating the new horizontal tension in the wire for each section, and calculating the point in the y co-ordinate that the mass hangs.

Wait, are you SUPPORTING the chain at point P or putting a load there? If you are putting a load there, which was how I understood it, I can't believe the two sides will be catenaries. Are you sure about that?

 

[Medgy]

Definitely putting a load there. It will add tension to the wire but the two sections of the curve should still form catenaries. It's basically like calling (Xp, Yp) a support, splitting the the curve in two, and adding some tension to the wire. It may push (Xc1, Yc1) outside the defined supports (Xc, Yc defines the co-ordinates of the lowest point of a catenary) but I see no reason why the two curves should not be catenaries. We're assuming a static environment.

 

[phinds]

Definitely putting a load there. It will add tension to the wire but the two sections of the curve should still form catenaries. It's basically like calling (Xp, Yp) a support, splitting the the curve in two, and adding some tension to the wire. It may push (Xc1, Yc1) outside the defined supports (Xc, Yc defines the co-ordinates of the lowest point of a catenary) but I see no reason why the two curves should not be catenaries. We're assuming a static environment.

I can't do any math to support it but I contend strongly that they will not be catenaries. Clearly you have a situation where neither side is a free-hanging cable.

Do what I do when faced with a puzzling math issue. Take it to the extreme. In this case the extreme is that you put a REALLY heavy load on at P. THEN what do you end up with? Two straight lines connecting the support points with the load point.

 

[Medgy]

Well what happens when you tension a single catenary (ie pull it from one side) with a gigantic force? You end up with a straight line. In my mind it's the same thing. You can apply tension to a catenary beyond the tension applied from just the weight per length and still have a catenary. I agree that at the point of application of the load, P, there will be a discontinuity. That's why I will need to split the curve in two at that point.

 

[phinds]

Well what happens when you tension a single catenary (ie pull it from one side) with a gigantic force? You end up with a straight line. In my mind it's the same thing. You can apply tension to a catenary beyond the tension applied from just the weight per length and still have a catenary. I agree that at the point of application of the load, P, there will be a discontinuity. That's why I will need to split the curve in two at that point.

Well, I still can't see that you'll have catenaries on both sides ... I don't think you have refuted my arguments at all. In any case, good luck with the problem and sorry I can't be any help.

 

[SteamKing]

Applying a point load to a cable suspended under its own weight will split the cable into two catenary arcs, one on each side of the point from where the load is suspended. Except for the point where this load is applied, the only force acting on the cable is its distributed weight, so that implies that the arc is that of a catenary, although it may be distorted somewhat due to the presence of the point load.

In any event, trying to work out the geometry of the resulting catenaries algebraically will probably be a complex and ultimately fruitless task. Since you want to solve for a given load suspended by a particular cable geometry, it's a task which can be analyzed from first principles and from which an iterative numerical solution can be obtained. Look at the statics of each cable arc: you know the horizontal tensions at the point load must sum to zero and the vertical components must sum to the weight of the cable and the point load.

 

[Medgy]

Well, I still can't see that you'll have catenaries on both sides ... I don't thing you have refuted my arguments at all. In any case, good luck with the problem and sorry I can't be any help.

I drew some diagrams to help with the explanation. Oh, an additional note that cables can't hold a moment.

 

[phinds]

Except for the point where this load is applied, the only force acting on the cable is its distributed weight, so that implies that the arc is that of a catenary, although it may be distorted somewhat due to the presence of the point load.

Why would the distortion not change it from being a catenary to not being a catenary? Why is my example of a really heavy weight not valid?

 

[Medgy]

Why would the distortion not change it from being a catenary to not being a catenary? Why is my example of a really heavy weight not valid?

I don't know how to explain it any better. Check out the first few pages of this resource: https://archive.org/stream/tensionsintrackc00ande#page/n5/mode/2up

 

[SteamKing]

Why would the distortion not change it from being a catenary to not being a catenary? Why is my example of a really heavy weight not valid?

Because how can you tell what a 'really heavy weight' is from what is just a 'regular heavy weight'?

In the limit, if enough tension is applied to an unbreakable line, a straight line will result. However, in the real world, there are no unbreakable lines, so the suspended cable is going to have some kind of sag, however small. The shape of the resulting curve can be approximated by a parabola under certain conditions, but the general shape is going to be some form of a catenary if the line has only its self-weight applied to it between the load and a point of support. Because of the non-linear math involved, an iterative solution to find the parameters of the catenaries is probably the most efficient.

If you google 'catenary with point load', you can find several references dealing with the analysis of this type of problem, which pops up in the design of elevated tramways and ski lifts, for instance.

 

[phinds]

Interesting. Thanks.