Iron bar bending force

  1. 1. The problem statement, all variables and given/known data

    Given an iron bar, of round cross-section, fixed by its extremities
    (bar measures 152 mm in length and 4.8 mm in diameter),
    a weight of 88.5 kgs is hung from its middle, bending it at least
    90 degrees.

    How much weight/force would be needed to bend another
    bar of the same material and size, but of hexagonal or square
    section?


    2. Relevant equations

    Torque = Force * R

    Sin(90°) * R


    3. The attempt at a solution

    Haven't got any idea. There should
    be something lacking in the text or
    more probably some calculus skills
    given as knows and hence omitted.

    Hope you could clear up this question,
    thank you in advance
     
  2. jcsd
  3. I'm not sure, but I think that you need a second moment of inertia over here, which is defined as

    [tex]I = \int r^2\, dA[/tex]

    where the are is perpendicular to the axis of bending. The bigger this moment of inertia is, the harder it is to bend or deform the beam, so you just use proportion to find what you need.
     
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