Tension force in catenary


by Engineering01
Tags: cable, catenary, cosh, mathematics
Engineering01
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#1
Dec20-12, 06:14 AM
P: 11
Hello all,

I have yet another mathematical quandary that is robbing me sleep, so I return to gather ideas.

This problem (fuelled from self-study) involves the equation of a catenary.

The profile of a catenary can be expressed as y(x) = T/w*(cosh(w/T*x) – 1). My inputs into this equation are:
  • y(L/2): Sag at mid-span
  • w: distributed load
  • x: position along the cable with respect to the origin. The origin is located at the mid-span of the sagging catenary profile.

Given these inputs I am looking to solve for the minimum tension on the cable, denoted T in the above equation. So far I have been using excel to solve this by “trial-and-error” using the solver add-on. My approach, thus far, has been to rearrange the above equation to the following form: y(L/2)*w/T = cosh(w/T*x) – 1. This has allowed me to set a cell with the RHS – LHS and hence use solver to find a solution.
Some interesting outcomes I have found so far are:
  • I have compared the results to a parabolic cable profile and seem to be achieving correct results (for the most part)
  • This method does not always work

For a given set of inputs (listed above) the solution may not converge because of the starting value used for T in the iteration. I have found that tweaking my initial guess for T may sometimes lead to a solution but I was hoping to be able to solve for the correct solution without helping solver all the time.

To the point: is there a fundamentally better way of being able to solve for the minimum cable tension in this equation? I am looking to do this explicitly in excel so this means by iteration (as I have attempted above) or solving for an expression for T (this route doesn’t seem too friendly).

Any thoughts on how to best manipulate this equation?

Cheers
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Studiot
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#2
Dec20-12, 06:21 AM
P: 5,462
It is usual to assume the weight of cable is negligable compared to the (uniformly) distributed load.

Can this be assumed in this case?
Studiot
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#3
Dec20-12, 06:34 AM
P: 5,462
Here is a start.

Since there are no horizontal loads the horizontal component of the tension (T) is constant (H).

It may be found by taking the origin for an xy coordinate system as the lowest point and considering moments at some point along the cable at (x,y) about the origin (O).

Then if s is the length along the cable curve.

[tex]T = H\frac{{ds}}{{dx}} = H\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} [/tex]

For a symmetrical cable, horizontal length L, the tension is a max at L/2 and equals


[tex]{T_{\max }} = H\sqrt {1 + \frac{{16{d^2}}}{{{L^2}}}} [/tex]

Engineering01
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#4
Dec20-12, 06:37 AM
P: 11

Tension force in catenary


Studiot,

I am building in the self weight into the Uniformly Distributed Load (UDL). You are 100% correct in stating that the self weight of the catenary is negligible in comparison to my applied load. So in essence I am creating a hypothetically dense cable. Here is one example:

w = 1kN/m
Distance between supports, L = 3m
y(L/2) = 20mm


Graph of results

Engineering01
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#5
Dec20-12, 06:48 AM
P: 11
Studiot,

Thanks for supplying those equations above - you have almost solved my problem. The only thing I am missing is how to find the horizontal component of force (H) in the cable seeing as I have two unknowns in the above equation. Is there another expression I can make use of?

Thanks for the help
Engineering01
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#6
Dec20-12, 04:09 PM
P: 11
I almost have this solved! What does 'd' stand for in the Tmax formula?

Thanks!
Studiot
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#7
Dec20-12, 04:34 PM
P: 5,462
D id the drop of the catenary from suspension point to lowest.

Sorry. I should have said.
haruspex
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#8
Dec20-12, 05:16 PM
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Quote Quote by Engineering01 View Post
You are 100% correct in stating that the self weight of the catenary is negligible in comparison to my applied load.
I'm confused. A catenary arises from a cable with uniform mass density, no applied load. If you have a uniform (with horizontal distance) applied load and the cable mass is negligible in comparison (as with a suspension bridge), the cable will adopt a parabola.
Another interesting case (with no applied load) is to make the cable thickness vary so that the cross sectional area is proportional to the tension. This gives a cycloid, as I recall.
PhanthomJay
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#9
Dec20-12, 06:01 PM
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As noted, the shape of the cable is parabolic, not that of a catenary, when uniformly loaded and the cable weight can be neglected. The minimum tension is the horizontal tension in the cable (H) which can be found by drawing a free body diagram through the low point of the span and around one end. Note that even if the cable is hanging just under its own weight and taking on the catenary shape, the catenary curve is very closely approximated by the parabola when the sag D is less than about 10 percent of the span; beyond that , the catenary and parabolic curves start to diverge.
Studiot
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#10
Dec20-12, 07:47 PM
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Just how do you apply a

If you have a uniform (with horizontal distance) applied load
to a cable?

Yes if you could it would take on the shape of a parabola and terminology in this thread so far has been somewhat loose.

If you suspend a deck witha a UDL below a cable via a finite number of suspenders the shape is neither a catenary, nor a parabola, though not far from either.

The formula I listed for Tmax is based upon a parabola (y = 4x2d/L2) and works for small d/L.

Other forms of load, differential supports, stiffening decks etc you would have to rework the length of curve expression and substitute the actual for the parabola
haruspex
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Dec20-12, 09:19 PM
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Quote Quote by Studiot View Post
Just how do you apply [that] to a cable?
Just trying to understand the requirement here. If the question is really about catenaries, why assume negligible self-weight?
Studiot
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#12
Dec21-12, 02:45 AM
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I have to admit I was thinking of bridges partly because I do and partly because a distributed 'load' was mentioned in th OP.

Of course for many, if not most, uses of load bearing suspended cables the weight of the cable is significant. Sometimes the cable weight is much greater than that of the load (catenary supports for communications cables) and may even be the only load (power cables).


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