1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Second moments of area of a simply supported regular I beam

  1. May 28, 2012 #1
    I’m having trouble getting my head around this example and solution that was given as part of a revision pack for an upcoming exam. Any explanation would be gratefully received.

    Question:

    A simply supported regular I beam is 90mm wide and 120mm high. The top and bottom flanges are 10mm thick, as is the web. It is 6m long and carries a UDL of 6KN/m over the length of its top flange.
    Determine the second moment of area Ixg about the horizontal axis which passes through the centroidal axis of the beam.


    Equations:

    Second moment of area of a rectangle about an axis through its centroidal axis, Igg= (bd^3)/12

    Second moment of area of a rectangle about its base, Izz= (bd^3)/3

    Parallel axes theorem, Iaa= Ig+A(y)^2 (I’m not sure how to put the – over the y using word)


    Solution:

    Total area A= 4x40x10+120x10= 2800mm^2 … easy enough

    Centroid y = [(2x10x40x5)+(2x10x40x115)+(120x10x60)]/2800 = 60mm

    Again this is easy enough and I can see the logic, but this is where I am struggling:

    From second moment of area of a rectangle about its base, Izz= (bd^3)/3 and parallel axes theorem Iaa= Ig+A(y)^2.

    Ixx= 1/3x40x10^3x2 + 1/3x10x120^3 + 2x [1/12 x 40x10^3 + 40x10x115^2] = 16.37x10^6 mm^4


    An explanation of the different sections of the line above would help me a lot. I’m confused with the use of Igg= (bd^3)/12 and Izz= (bd^3)/3.


    Ig= 16.37x10^6 – 2800 x 60^2 = 6.29x10^6 mm^4… so Ixx is in fact Iaa?


    Thanks for taking the time to read this and I hope that question is clear.

    Cheers Steve.
     
  2. jcsd
  3. May 28, 2012 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    When using the parallel axis theorem for determining the 2nd moment of area with respect to the axis passing horizontally through its cg, use the I of the rectangles about their own cg's. These are all of the form bh^3/12. To this number, you must add the products of the individual rectangle areas times their distance^2 from their own cg to the member cg
     
  4. May 30, 2012 #3
    Thanks for your reply.

    I've taken on board what you have said and I think I'm getting it now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Second moments of area of a simply supported regular I beam
  1. Simply supported beam (Replies: 4)

Loading...