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What should I do?

This is a discussion on What should I do? within the Suspension and Handling forums, part of the General Help category; I'm not doubting Jon..... That is based on if you have equal mass of the material and equal quality/type of ...

  1. #21
    Spaz is My Mentor SMWS6TA's Avatar
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    I'm not doubting Jon.....


    That is based on if you have equal mass of the material and equal quality/type of material. Correct Jon?

  2. #22
    Senior Member kenro23's Avatar
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    Im not doubting either. I am just curious what they were
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    Moderator 35th-ANV-SS's Avatar
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    Here is my thought process...

    The area of a tube is the area of the outside diameter minus the area of the hole of the tube. So, if you have a 2" diameter tube with a 1" hole in it (1/2" thickness), the area would be:

    Outside area = PI * Radius^2 = 3.14 * 1"^2 = 3.14 inches^2
    Hole area = PI * Radius^2 = 3.14 * 0.5"^2 = 0.785 inches^2
    Overall area = 3.14"^2 minus 0.785 inches^2 = 2.355 inches^2

    Now let's look at the same size in a box (square) in a 2" by 2" that is 1/2" thick.
    Area of outside of square = length X width = 2" X 2" = 4 inches^2
    Area of the inside of the square (empty part) = 1" X 1" = 1 inch^2 (it's only 1" because you subtract 1/2" from each side for the thickness).
    Overall area = 4 inches^2 minus 1 inch^2 = 3 inches^2

    So you are now comparing 3 inches^2 of area versus 2.355 inches^2 of area.

    You can also consider the moment of inertia of each piece, tubular versus a square/box and you will find that the moment of inertia of the box is higher.

    Bending stress = Mc/I, where M = moment, c = distance from the neutral axis to the extreme fiber, and I = moment of inertia. So, the higher the value of I, the lower the bending stress.

  4. #24
    Senior Member kenro23's Avatar
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    Wow

  5. #25
    Member derekbuckner1's Avatar
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    ^ what he said. That was impressive.

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