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Trouble calculating the second moment of area

  1. Mar 31, 2008 #1
    I have partial circular sector, in the shape of a smile with the right corner of the smile at the origin, and the the 2 upper most points of the smile coincident with the x axis (see jpeg attachment below for pictorial)

    This is a cross section of a 91" piece that will be rigidly supported on each end.

    I'm trying to calculate the 2nd moment of area for it with the goal of calculating the deflection of it.

    I tried using this equation:

    Second moment of area = Ix= 1/4(Theta+1/2sin(2*Theta))(R^4-r^4)

    where theta in this case = 1/2 the angle of the total piece = 15.57 degrees or 0.272 radians, and R = 7.116181102in and r = 7.096496063in (this is a very thin, narrow piece at around .020" thick only about 4" wide along x axis)

    However, when I stuck the results of this equation into my deflection equation, (Weight* Length^3)/(384*E (aluminum)* Ix)

    Where my weight is .72 lbs, length is 91", E of aluminum = 10.3* 10^6, and Ix=3.75 in^4,

    the deflection comes out to only approximately .00003" which seems way too small.

    When I use similar numbers and calculate the deflection of a rectangular beam with base of 4in, height of .020 inches, I get a huge deflection (Ix = 1/12 * b * h^3), = 54 inches. This seems too big but maybe this just means that the piece will not support it's own weight?

    I can't figure out what I'm doing wrong, but I can't believe that adding the radius shown in the jpeg to this almost 8 foot piece of aluminum (91 inches) would improve the deflection of the piece by that much.

    Help! This is not just theoretical help for school, I am trying to design a new solar product, and we will spend $$$ on machining of this product so it has to be right before we do the tooling.

    Am I doing something wrong with the rectangular beam calculation? If not, then there has to be something wrong with the Second moment of my cross section?

    Thanks for any help you can give me.
     

    Attached Files:

  2. jcsd
  3. Apr 15, 2008 #2
    If this piece is rigidly supported at the ends won't it be holding the shape that you give to it when you rigidly support it? Are you going to load it beyond its self weight?

    Chris
     
  4. Apr 15, 2008 #3
    It will hold it's shape on the ends, but it is 91" long so will deflect in the middle the most, just like an I-beam would if you held it on the ends, it would still deflect in the middle.
     
  5. Apr 15, 2008 #4
    I believe you will have to treat this as a cable with a uniformly distributed weight along it's length rather than as a beam. The curve involved would be a catenary. The shape (and hence the deflection) will depend on the exact distance between the fixing points and the exact length of strip.

    Chris
     
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