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Beam bending of an angled flat plate

  1. Jul 30, 2014 #1
    Hello everyone,
    I am wondering how to properly go about calculating the stresses caused by bending on an angled flat plate. Two approaches seem valid but they give much different results:
    -Calculate the new area moments of inertia for the angled flat plate (new Ixx, ymax).
    -Resolve the moments into parallel and perpendicular components, then add the calculated stresses from the principal area moments of intertia (Ixx, Iyy).

    I wrote a short script to compare the resulting stresses from these two methods and they give answers that are different by one or two orders of magnitude. The first method gives lower stresses (more stiff).
    I cannot reason how which method is right. I do believe the second method is right, but I cannot explain why the first method is wrong. I know I am missing something with the first method.
     
    Last edited: Jul 30, 2014
  2. jcsd
  3. Aug 3, 2014 #2
    Here's how I did the first method, maybe they are similar as far as what you did.

    t=thickness
    theta=angle at which it is inclined
    L=width of beam

    [itex]I_{yy} = \int y^{2}\frac{t}{sin(\theta)}dy = y^{3}\frac{t}{3sin(\theta)}[/itex]
    [itex]y_{max} = Lsin(\theta)/2[/itex]

    I was very careful deriving this, so I hope this helps a little bit as far as checking to see if you got the same answer. I'm not sure how to do the second part. I'm not sure how to resolve bending moments into parallel and perpendicular components. As it turns out, the x direction moment area is also [itex]I_{xx} = \int x^{2}\frac{t}{sin(\theta)}dx = x^{3}\frac{t}{3sin(\theta)}[/itex], so at least that part is easy enough... However, note, you have to use [itex]x_{max}[/itex] for the x direction component, and that will come out to be
    [itex]x_{max} = Lcos(\theta)/2[/itex]

    Again, can't really help on separating the components, never had to do that before...

    DISCLAIMER: Something doesn't quite seem right about this solution, though I've carefully checked it several times. If you see anything that looks wrong to you let me know and I'll look at it again.
     
    Last edited: Aug 3, 2014
  4. Aug 24, 2014 #3
    Oh, hey, I have to add a correction. I did not use the centroid as the origin, but the corner (which is wrong). You can adjust for that by using the parallel axis theorem (subtracting A*∆x^2 or A*∆y^2 for Ixx or Iyy respectively, where the centroid is (∆x,∆y) ).
     
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