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Homework Help: Arch Bridge Compression Calculations - Help

  1. Nov 19, 2013 #1
    Hi all, basically we're building a model bridge (30 grams) for a mechanical engineering project out of balsa wood and kevlar (for the strings). I've attatched a pdf file of the bridge design, although it doesn't show the strings.

    We are wanting to do some basic load compression calculations on the arches and we can't find any relevant or simplistic equations.
    If someone could point us in the correct direction, that would be great.


    Attached Files:

  2. jcsd
  3. Nov 23, 2013 #2
  4. Nov 23, 2013 #3


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    Lambro6: The drawing is currently missing a lot of dimensions, and cross-sectional dimensions, joint details, and, like you said, the string locations, and also the applied load, and external constraints. It currently looks like the arch is cut into three segments, then glued to member 5. But I cannot really tell. How is member 1 constrained? Is member 1 free to rock (which would be a pinned support), and free to slide horizontally (which would be a roller support)?
    Last edited: Nov 23, 2013
  5. Nov 24, 2013 #4
    I can give you all the dimensions which you've stated as missing, but we're really looking for a simple analysis of the load that will be put on the arches (hopefully achieving around 20kg - as the test for our bridge is one to destruction).

    The 2 arches are simply 2 long beams of balsa wood that have been bent to form an arch. They are then held apart by the 4 beams going across. The load will be placed on the platform which will be attached to the arches by strings. There is also a string which is attached to either end of both arches, pulling the bridge together.

    I have pictures of the test bridge if it would help?
  6. Nov 24, 2013 #5


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    In general, stress analysis of curved beams is more complicated mathematically than for straight beams. There are no simple formulas like you find for straight beams, and you have to go back to first principles. Rigid arches supporting a span using suspenders puts the problem further out there.

    http://courses.washington.edu/me354a/Curved Beams.pdf

    Not being a bridge engineer, I suspect most of these problems are solved nowadays using numerical methods (like FEA).
  7. Nov 25, 2013 #6


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    Lambro6: (1) How do you create the shape of your arch? Do you just bend member 2, when it is dry, and install the horizontal string?

    (2) What are the cross-sectional dimensions of the arch member (member 2)? Is the arch member a 10.0 mm by 10.0 mm square cross section?
    Last edited: Nov 25, 2013
  8. Nov 25, 2013 #7
    (1) by placing the arch in hot water for an hour then bending it and leaving it to dry in the bent form. And yes, then wrapping and knotting the string around the arches.

    (2) yes it is a 9x9 cross section spanning the distance of approximately 520mm
  9. Nov 25, 2013 #8


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    Lambro6: Your bridge currently looks statically indeterminate, I think. Therefore, there is probably not a simplistic equation. It would probably require a structural analysis.

    Your arch member (member 2) is a 9.0 mm square cross section (b = 9.0 mm). By scaling your drawing, I assumed the height of your arch (from the ground surface to the member 2 centerline) is a2 = 126.5 mm, and I assumed half of the arch horizontal length (from the bridge vertical centerline to where the member 2 centerline intersects the ground surface) is a1 = 275.0 mm. I assumed the angle subtended by your arch, from the arch center to the arch end, is theta = 49.4 deg. Therefore, I assumed the arch radius of curvature is rho = a1/sin(theta) = 362.2 mm. I currently assumed your balsa bending ultimate strength is Stu = 20.5 MPa (?). I conservatively assumed all of your applied load (P1 = 196.2 N) is applied to the two midspan vertical strings. I assumed positive bending moment at the arch midspan causes tensile stress in the arch bottom fiber.

    Therefore, for this particular arch, the arch maximum bending stress occurs at the arch midspan (x = 275.0 mm). The arch midspan axial stress, plus top fiber bending stress, due to applied load P1 is, sigma1 = -0.984 - 28.03 = -29.01 MPa. Therefore, sigma1/P1 = -0.14786 MPa/N.

    We want, -Stu = (sigma1/P1)*P2. Therefore, -20.5 = -0.14786*P2. Solving for P2 currently gives, P2 = 138.6 N. Therefore, if all of the above assumptions are correct, this currently seems to indicate the bridge could potentially support a midspan applied mass of P2/g = 14.1 kg (?). These results have not been checked.
    Last edited: Nov 25, 2013
  10. Dec 7, 2013 #9
    Thanks very much for the effort you've put in here, I appreciate it.

    Can I ask, what is the formula(s) you used to put all those numbers into?
  11. Dec 8, 2013 #10
    There is a significant difference between a beam bent into an arch shape, and simply supported at its ends (on the one hand) and an arch whose abutments are prevented from moving horizontally relative to each other by heavy bases or a tie between them (on the other hand). In either case, the bending moment diagram is a key to understanding the problem. Can you draw that?
  12. Dec 10, 2013 #11
    No I'm not sure how to draw a bending moment diagram for such a shape.

    We've only covered simple beams and truss structures.
  13. Dec 10, 2013 #12
    Are you able to prevent the supports from moving apart, either with a tie connecting the feet, or by heavy bases that will overcome friction? If not, and in any case, draw the bending moment diagram as if it were a simple beam.
  14. Dec 10, 2013 #13
    Yes there is Kevlar holding the arches together from either end so they can't spread apart.
    If you look at my other thread I've explained it slightly better.

  15. Dec 11, 2013 #14
    If the load were a central point load W, then, on a simple beam the M diagram would be a triangle with peak M WL/4 at the centre. If the load is distributed then it would be a bit reduced. However the important part is that you can graphically superimpose the shape of the arch on the triangle (or trapezoid, if two loads, and much of the bending moment is cancelled out. The M diagram then has zero moments at roughly the 1/3 points and significantly reduced moments elsewhere. Another useful way to look at it is that, to some scale, the M diagram (drawn on the same side of the axis as the arch is itself) represents a line of thrust in the arch. Deviations of the line of thrust from the arch centreline represent moment there. Even if you could do these calculations, the results would be affected by the prestress you have built in from constructing the arch shape. I think the better approach is to just make a model and test it, observing carefully where and how it fails. ' Line of thrust' is a worthwhile look up in google images. It may help, but I haven't had time to check that.
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