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Cables help please!

  1. Dec 1, 2007 #1
    [SOLVED] Cables help please!

    Hi, I don't think this is really a 'homework' question but more of a theory related question. I am currently a self-learner on engineering statics and at the moment I am reading up on cable theory and analysis, in particular the catenary problem. I am not used to physically interpreting mathematical equations/data and was wondering if someone could be of help?

    Okay so basically, lets consider a perfect inextensible and flexible cable of length 2L, suspended by its ends from two supports B and B' a distance 2l apart and at the same height. We define c = T_o/mu (sorry I am not a use of Latex) and the equation for the catenary shape has been derived to be y = c cosh (x/c) after choosing A to be our lowest point of the cable as co-ordinate (0,c).

    Also by using the arc length formula we find the arc length s (from A to B) is s = c sinh (x/c) => L = c sinh (l/c) since from A to B = L and x = l

    What I don't understand is the next part where we have non-dimensional variables.

    [​IMG]

    [​IMG]

    could someone explain to me:

    1) What is the purpose of non-dimensionalizing?

    2) Why on the diagram is there an asymptotic line x = 1? Why is L always greater than 1?

    3) Why is it a reasonable physical explanation that L > l?

    Sorry for the silly questions but I've never really had the right set of mind in understanding physical interpreations and math results.

    Thanks all in advance! And as I am reading from a text and notes, I have just summarised the theory and results. If anything is unclear please let me know and I will try to explain more fully.
     
  2. jcsd
  3. Dec 1, 2007 #2

    PhanthomJay

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    I assume that as a self learner in statics that you are at the tail end of your studies, since the catenary curve with its hyperbolic functions is certainly not a place to start!
    Non dimensionalizing (by dividing by l) just simplifies the equation. You are plotting L*(not L) versus c*. 2L is the total length of the cable as measured along the catenary curve, whereas 2l is the horizontal distance between its supports. Since the cable is flexible, it therefore has sag (shown as h on your sketch), so it is reasonable to note that the cable length 2L must always be greater than the distance 2l between its supports, that is, L is always greater than l, thus L*is always greater than 1. The curve shows that as the sag between supports increase (as for example,when the cable is fed through its supports to increase L while l remains constant), then the value of c* (which is a function of the tension in the cable) decreases. The value of c* becomes infinite at L*=1, since at this point, the cable length is equal to the distance between supports, implying no sag at all in the wire, and hence infinite tension (of course, the wire would break before reaching infinite load)!
     
    Last edited: Dec 2, 2007
  4. Dec 2, 2007 #3
    Hi, many thanks for your great response! It was very helpful and I am understanding the underlying idea. Yes I am nearing the end of my studies and structural mechanics/engineering has always been an interest to me so I'm just teaching myself through textbooks and notes from one of my lecturers :smile:

    I hope you don't mind me asking another question relating to the catenary problem, this time it is computational results relating to the sag :redface:

    If you could help me out that would be excellent. Many many thanks in advance!

    Having derived the equation for the sag, we want to make observations on the relationship between the sag and L.

    So considering two cases, L -> 1+ (as L gets small) and L-> infinity (as L gets large), deduce the behaviour of sag h.

    I've found out that as L gets large (tends to infinity) h ~ L (behaves like L) and as L -> 1+, h behaves like a parabolic function.

    On the following graph the results are shown, with exact solution of h (solid red line), h as L -> infinity (green line) and h as L-> 1+ (blue line).

    [​IMG]

    Again I am not sure how to interpret the data in relation to cables..:confused:
     
  5. Dec 2, 2007 #4

    PhanthomJay

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    Excellent observation. The catenary equation, in spite of the beauty of its curve, leaves me cold. Fortunately, for large L/h or l/h ratios, that is, where the span lengths/cable lengths are much greater than the sags, the catenary curve can be approximated by a parabolic curve, where h= wl^2/8T_o, where h is the sag, w is the unit weight of the cable, and T_o is the horizintal cable tension at its low point. You can see that if the tension remains constant and you double the span, the sag increases by 4 (the sag varies as the square of the span). For small span/sag ratios, the approximatation is no longer valid (the cable weight starts controlling over the cable tension) and you're stuck having to use the complexities of the hyperbolic functions.
     
  6. Dec 2, 2007 #5
    Hmm okay, to be honest I'm still slightly overwhelmed! :redface: so the illustration of the exact solution of h on the graph, is it an incoporation of both approximations? regarding the exact solution, why is it that it approximates so close to the parabolic representation but it starts to converge more and more towards the straight line approximation? :confused: (sorry if you have already answered my question, im still slightly confused!)
     
  7. Dec 2, 2007 #6

    PhanthomJay

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    No, the solid red line for the sag as a function of span length is the exact catenary solution. The blue prabolic curve approximates the exact solution when L* is close to 1 (large span/sag ratios, where l is only a few percent greater than L). The green line (h=L) approximates the exact curve only when span/sag ratios are small. For example, when the sag is say less than a few percent of the span, you can approximate the catenary with the parabolic curve; when the sags are huge in comparison to the span, only then can you approximate the curve with the straight line. Very difficult concept, for sure.
     
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