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Suspension bridge problem

  1. Sep 2, 2015 #1
    1. The problem statement, all variables and given/known data

    A simple suspension bridge consists of a uniform plane horizontal rectangular roadway of mass M, freely supported by two sets of four equally spaced light vertical cables. Each set of four cables lies in the same vertical plane and is attached to one edge of the roadway. These cables are in turn attached to a single light cable ABCDEF as shown below. A, B, C are on the same horizontal level as F, E, D respectively.

    untitled.GIF

    A and F are 8 m and C and D are 2 m above the level of the roadway which has length 20 m. AF = 20 m. Given that the tensions in all vertical cables are the same, show that the tension in CD is Mg/4 and find the tensions in each of the parts AB, BC, DE, EF of the connecting cable.

    2. Relevant equations



    3. The attempt at a solution

    resolving vertically: Mg = 4T (T = tension in all vertical cables).
    Actually, I'm not sure of this. What about the reactions at A and F?

    Let tension in BC = T1
    Let a = acute angle BC makes with CD

    tension in CD = T1 cos a
    now I'm stuck. How do I find cos of a?

    any help greatly appreciated,
    Jimi
     
  2. jcsd
  3. Sep 2, 2015 #2

    andrevdh

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    You can get a with the 4 m and the 2 m
     
  4. Sep 2, 2015 #3

    andrewkirk

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    Not unless it says somewhere that the height of B and E above the road is 4m. So far as I can see, it doesn't give any heights for B or E. that seems like an omission in the problem specification, unless it may be that the angle a does not affect the answer because of some cancellation phenomenon.

    Hendrix are you sure the problem does not give heights for B or E?

    Edit: Why not try writing out the equations for all the forces in terms of a as an unknown, then see if a cancels out in the final formulas.
     
  5. Sep 2, 2015 #4
    andrevh - I don't see how since I don't know how high the vertical cable is from the road to B.
     
  6. Sep 2, 2015 #5
    andrewkirk - no. I've checked the problem and this is all there is. Was thinking maybe there's a way of using the height of 8 m but I can't see how.
     
  7. Sep 2, 2015 #6

    andrewkirk

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    The problem statement is definitely missing something, because you are asked to find out the tension in BC, which is ##\frac{Mg}{4\sin a}##. There is no possibility for the ##a## to cancel out in that formula, even if it cancels out in some of the other tensions (which I doubt it will).

    I suggest you ask your lecturer whether:

    (1) there is a piece of info missing from the problem statement, such as the height of B; or

    (2) you are expected to give the answer in terms of some unknown quantity such as ##a##.
     
  8. Sep 2, 2015 #7

    andrevdh

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    The vector sum of the tensions T and TCD need to cancel TBC - draw a vector diagram.
     
  9. Sep 2, 2015 #8

    andrevdh

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    With multiple equations you can solve for more than one unknown quantity.
     
  10. Sep 2, 2015 #9
    Try cutting the cable between C and D and summing torques on half the "U" cable.
     
  11. Sep 2, 2015 #10

    andrewkirk

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    I just realised that the 2m distance from C and D to the roadway is irrelevant to the problem, given that the usual assumption of massless cables is used. It doesn't affect any of the tensions. Why would they give a useless piece of data, while omitting a necessary piece? Because the problem-setter made the mistake of thinking that the datum given would enable us to work out the angle a. I give odds 20 to 1 that the 2m was meant to give the angle a, and the problem-setter didn't realise that it didn't (even uni lecturers make mistakes sometimes:wink:).

    Given that observation, I suggest you begin your answer as follows.

    'The information needed to determine the angle of BC to the horizontal, and hence the tension in cable section BC, is missing from the problem statement, yet an unusable piece of information is given that the distance from B to the roadway is 2m. Hence I will instead assume that the vertical height of B above C is 2m. ........
     
  12. Sep 2, 2015 #11
    I was able to solve the problem completely with the data given. And yes, the 2m distance is irrelevant.
     
  13. Sep 2, 2015 #12

    andrewkirk

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    Really? how?

    If B is 1m above C the tension in BC will have to be greater than if it is 2.5m above it. Yet it seems to me that a bridge conforming to the problem statement can be built with either height of B.
     
  14. Sep 2, 2015 #13
    Start with Post #9.
     
  15. Sep 2, 2015 #14

    andrewkirk

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    • Poster has been reminded that the student must do the bulk of the work...
    I don't understand post 9, as torques don't seem to me to be a natural way to approach the problem.

    However since my last post I've realised that the problem is solvable as stated, because the forces in the cable have to align with the cables. So we solve it as follows.

    There are Five unknown forces:
    ABv : the vertical component of tension in AB
    ABh : the horizontal component of tension in AB
    BCv : the vertical component of tension in BC
    BCh : the horizontal component of tension in BC
    CD : the tension in CD, which is horizontal

    We have the following five equations:

    ABv = Mg/4 (each end has to support one quarter of the bridge)
    ABh=BCh (horizontal equilibrium at B)
    ABv=BCv+Mg/8 (vertical equilibrium at B)
    BCh=CD (horizontal equilibrium at C)
    BCv=Mg/8 (vertical equilibrium at C)

    [Note: corrections were made to 1st, 3rd & 5th eqns per insightful's post below]

    So we can solve the equations to find the vertical and horizontal components of tension in each cable section.
    We then find the the total tension in each segment by Pythagoras.

    If we want the angles, we use the fact that the cables must align with the force in them, and calculate the angle from horizontal as
    arctan(vertical compt of tension / horiz compt of tension).

    What was throwing me was that we can build the bridge with a range of different angles for AB and BC. What I didn't realise is that doing that with anything other than the angles that fall out of the above equations will cause the points B and E to move laterally as well as vertically, so that the cables going down from B and E are no longer vertical. Since the problem statement requires that they are vertical, there is only one possible set of angles that works.

    An interesting practial realisation from this, for me, is that, when the load changes on a suspension bridge - eg between light and heavy traffic - there will be slight lateral displacement in the (almost) vertical cables.
     
    Last edited: Sep 2, 2015
  16. Sep 2, 2015 #15
    Are you forgetting that there are two cable systems supporting the bridge (one on each side of the roadbed)?
     
  17. Sep 2, 2015 #16

    andrewkirk

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    Good point. Yes I was. Anything involving Mg needs to be divided by another factor of 2. I'll make the corrections in the post
     
  18. Sep 2, 2015 #17
    So, how do you solve for CD?
     
  19. Sep 2, 2015 #18

    andrewkirk

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    To compensate for the 1st eqn being the sum of the 3rd and the 5th, we use the extra equation that I didn't write down :

    (ABv+BCv)/(ABh+BCh)=(8-2)/(4+4)

    So it seems we do need to use the given 2m after all.
     
    Last edited: Sep 2, 2015
  20. Sep 3, 2015 #19

    NascentOxygen

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    Consider, say, the left half of the suspension cable, and take sum of moments about A.
     
  21. Sep 3, 2015 #20
    Exactly!
     
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