Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Second moment of inertia for a bent rectangle

  1. Mar 20, 2017 #1
    I am currently working with a beam with the following cross-section:
    It consist of three bended sections with the following parameters, alpha = 90 degrees, Thickness = 4 mm, Radius = 50.59 mm.
    The top section consist of a small triangle and a rectangle. the triangle have a width = 4 mm and height = 2.81 mm. The rectangle have width = 4 mm and height = 19.60 mm.

    The area of the profile is given by the following equation, and should be approximately 1000 mm2:

    I need to find the deflection of this beam, and thereby the second moment of inertia. I have allready found this to be 4273323.41 mm4 in Section Properties in Solidworks, but this need to be substantiated by hand calculations. Does anybody know how to calculate the second moment of inertia for such a geometry?

    Best regards,
    Last edited: Mar 20, 2017
  2. jcsd
  3. Mar 20, 2017 #2
    You want the second moment with respect to what axis?

    Which ever axis it is, just apply the definition and carry out the integration.
  4. Mar 20, 2017 #3
    Thank you for answering. I would like to find Iyy, which in this case mean in vertical direction.
    Which definition do you mean?
  5. Mar 20, 2017 #4
    Your figure is unclear. Is the sinuous figure at the top truly a cross section, or is it a plan view? Where is the y-axis in that view? Which way is the loading applied?

    You need a three dimensional presentation of your problem to make clear what the situation really is.
    There are definitions online and elsewhere for the second moment of an area expressed in terms of integration.
  6. Mar 20, 2017 #5
    Both figures are captured from the front plane, but shows the cross-section of the model. The x and y-axis is shown in both figures, and y is in vertical direction. The loading is applied in negative y-direction.

    I have the 3D model, and can extract all information from SolidWorks, but these values need to be substantiated by calculations. But the second (area) moment of inertia only depends on the area, why would I need a 3D presentation?

    Do you have any links, or formulas I can use to solve the problem?
  7. Mar 20, 2017 #6
    If the load is applied in the negative y direction as you say, this will tend to straighten the "crooked stick." This straightening action will involve bending (actually reducing the radius of curvature) in each of the arcs. Bending of this sort depends upon the second moment perpendicular to the x-y plane, Izz. You have not shown us what that section looks like.
  8. Mar 20, 2017 #7
    The model shown in the figure is extruded 11 meters in the z-direction. In perspective, the height of the beam in y-direction is approximately 220 mm. So it is very long compared to the cross-section. I think I might have been a little unclear in the problem desciption. I want to analyse the deflection when the beam is fixed in both ends, and subjected to a uniform load which represents its own weight

    I've been looking into a few other beam profiles, i.e, hollow tube, hollow triangle and T-profile, all with approximately the same cross-section area and the same length in z-direction. Assuming the same coordinate system as for this model, the deflection off the beam could be decided by only looking at Iyy for all these other beams. Why would it be any different for this model?
  9. Mar 20, 2017 #8
    With the z dimension at 11 m, it sounds like a plate bending problem, not a beam problem.

    However, I seem to be unable to communicate with you clearly, so I'm going to drop out and let someone else try.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted