Area Moment Of Inertia for an Inclined Beam

Hi,
Can anyone help me with an area moment of inertia problem that i have?
I'm idealising a cross-section of an aircraft longeron and one of the
flange of the longeron is inclined at at an angle. While there is a
formula to find the area moment of inertia of a inclined beam (I'm
idealise that inclined flange into an inclined beam), there is nothing
about which distance to use when i have to use parallel axis theorem to
find the total area moment of interia for the whole cross-section.
A fellow engineer told me to use the distance from the centroid of the
inclined beam e.g. treat it like a normal vertical or horizontal
element, but it is correct? I think that for one axis, the assumption
is approximately correct but not correct for the other axis. Its a bit
hard to describe my problem in words so if you need the diagram to
understand my problem, please drop me an email at
arrow snipped-for-privacy@hotmail.com.
Thanks a million!
Ethan Ng
Reply to
F-5 Techie
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You could of course write equation for the various elements and integrate them to an exact answer. But I suspect you are looking for a bit of an approximation. As such, depending upon the size of the local areas with respect to the overall beams, you can ignore the errors associated with treating the areas as descrete points.
However, all that is required is to calculate the correct moments of inertia for each element within its own coordinate systems (orthogonal to useful axes) and then rotate the results into a universal system which is parallel with the desired axis. Then use the parallel axis theorem to translate those results to the universal coordinates and sum them.
A third methodology would merely to do what every solid modeling software does. Divide your areas into small regular shapes of size significantly smaller than the total area. Calculate the contribution of each small area to the overall moment of inertia based upon its descrete X-Y location and merely ignore the small inertia of the area itself.
My fee is a bit lower than that.
Reply to
oconnell

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