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Octagonal Cross Section with Parabolically Increasing Span

  1. Apr 21, 2013 #1
    Octagonal Cross Section with Parabolically Increasing "Span"

    Hi all, could some one tell me what I am doing wrong in my analysis in the attached file?

    Its a parabolical flare column with an octagonal cross section. The effect of the parabolic flare simply stretches the octagon so that the problem can be reduced to rectangle with parabolic increasing length with a constant height with an added term which adds the edge areas for each unit length. I feel confident in my analysis but it's giving me incorrect answers. I keep checking over and over and can't find an issue with my analysis.

    When I plug in test values in my final formula, I get 50.7 cubic yards but I know for sure that the answer is 35.7 cubic yards. In the attached files here are the variable values that I am using:

    b=5.5 ft
    s=11 ft
    p=16.5 ft
    l= 12 ft

    A=-2.5
    B=26.625
    C=-54.313


    I really cannot see the error. If anyone can help with this it'd be much appreciated.
     

    Attached Files:

  2. jcsd
  3. Apr 21, 2013 #2

    SteamKing

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    It's not clear from your calculations what l = 12 feet represents.

    The equation for the parabola is also uncertain. Are the coefficients of the parabola
    determined so that if x = 0, y = b/2 and if x = 16.5, y = s/2? This doesn't appear to be the case
    with the values of A, B, and C given in the OP.
     
  4. Apr 21, 2013 #3
    Sorry let me clarify.

    The variables are as follows:

    b=width of column at base
    s=width of column at the top
    p=length of column defined with x parabolically
    l=length of column defined with x constantly

    Note that the widths are divided by two since the y-axis is placed in the center of the column.


    The parabolic section has the equation of the parabola defined such that the x-axis lies on the top of the constant cross section segment. The conditions used to find the parabola were:

    x=2.75 ft, y=0

    x=3.5 ft, y=8.25 ft

    x=5.5 ft, y=16.5 ft

    Solving for the coefficients gives A=-2.5, B=26.62, C=-54.313

    This should be correct as the parabola should be:

    *concave down (a is negative, check)
    *intersecting the as defined x-axis with the left edge at a positive x-value meaning that the y-intercept should be negative (c is negative, check)

    Additionally creating a table in excel with that formula and specifying the values and creating a graph yield the proper shape so I'm positive that this portion is correct. Something is wrong in the integration I think. I'm thinking that since the parabola defines the outside edge, that my "x" listed in the drawing isn't really the x to integrate with but that x is actually, x'=x-b/2. I'm going to work that out and see if it comes out correctly.

    EDIT: Unfortunately this doesn't work either. I really have no idea what I'm doing wrong here.
     
    Last edited: Apr 21, 2013
  5. Apr 21, 2013 #4

    SteamKing

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    You've done a lot of calculations with the parabola and its equation. I decided to check your calculation using a quick numerical approach. I calculate a total volume of about 21.76 cubic yards, which gives you a third number to deal with, unfortunately.

    Since you know the dimensions of the cross sections of the column at top, bottom, and half way up the height, you can calculate the areas of these cross sections. Once these three areas are calculated, the volume of the column can be determined by using Simpson's First Rule of numerical integration.

    If the area at the bottom is A1, the area half way up is A2, and the area at the top is A3, then the volume = (A1 + 4*A2 + A3) * (Height / 6). This procedure, I feel, is much more rapid than trying to figure out the calculus and is less error-prone.

    My calculations gave the following areas in sq. feet, if you would like to check:
    A1 = 25.06
    A2 = 33.31
    A3 = 55.31
     
  6. Apr 22, 2013 #5
    Thank you Steamking for that, I hadn't heard of that numerical method and it does produce an appropriate answer. Further more it's also programmable (part of the purpose of me doing this). However that aside, I'm incredibly frustrated at this point that I haven't been able to derive this formula satisfactorily. For now I'm going to apply the Simpson's formula in my program but I would really like some help in pointing what I am doing wrong. I feel like this should have been a simple calculus problem but it seems to be escaping me. And my frustration only grows as my stack of rewritten scratch papers does. AND THAT RELATION IS NOT LINEAR.
     
  7. Apr 24, 2013 #6

    haruspex

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    From the diagrams, it looks like x is the additional octagon length, compared with the base. So x should start at 0, no?
     
  8. Apr 26, 2013 #7
    I realized after going over my diagrams that I was defining x incorrectly for the integration. It was properly defined in the cross section views but the equation I was using for x in the area formula was that for the parabolic profile of the OUTSIDE wall which is actually shifted by 1/2*b away from the actual x that I am trying to use to find volume. I'll recalculate my entire formula over again to double check, but preliminary calcs show I am getting around 30.8 cubic yards which is much closer than before and closer to other agreed values. I'll post more when I finish.
     
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