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Calculating I = 2nd Moment Of Area (m4) on C section beam

  1. Feb 12, 2013 #1
    https://skydrive.live.com/redir?resid=22F7BBABD7DEE83B!136&authkey=!AP6uQQ9mR04UuAE

    Can anyone please help.

    I have calculated the area and Neutral Axis about xx and yy of a C section beam which I know is correct (see above link)

    I'm have difficultly calculating the 2nd moment of area about xx & yy because the shape is not symmetrical, I have shown my workings but I know its incorrect. Can anyone help.

    I am also unsure how to calculate the Radius of curvature about the neutral axis xx & yy

    I would appreciate if you could show me step by step to give me a better understanding.

    Thanks
    hmk999
     
  2. jcsd
  3. Feb 12, 2013 #2

    SteamKing

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    Your calculation shows the inertia calculated about the horizontal axis.

    There appears to be an extra 0 in your final result.

    I did not find a similar inertia calculation about the vertical axis.

    I am attaching a tabular form version of your inertia calculation.
    This form is more compact, easier to check, and able to handle calculations
    involving a large number of parts. It is superior to what most textbooks
    show for similar calculations. It is also readily adapted to calculation
    with a spreadsheet like Excel.
     

    Attached Files:

  4. Feb 12, 2013 #3
    Thank you for your reply, much appreciated.

    I was unsure where to start about the vertical axis.

    Could you please attach with the vertical answer as well?

    I can then try and break it down to understand for future moment of inertia calculations I have to do.

    Thanks again
    hmk999
     
  5. Feb 12, 2013 #4

    SteamKing

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    It's a similar calculation to the one for the horizontal axis.

    I recommend you use the left side of the channel as the reference.

    Post your work and I'll be glad to check it.
     
  6. Feb 13, 2013 #5
    Hi, can you explain how you calculated the I values?

    Part A = 12500
    Part B = 20747573
    Part C = 8333

    Cant see how there calculated.

    Secondly, in part of the calculation INA = ΣI0 + ΣA*y2 - A*y2
    INA = 186116027 - 109396827 INA = 76719200 mm4

    I get 5420*142.07squared = 109396656 not 109396827.

    Cant see what I'm doing wrong.

    Can you please explain

    Thanks
    hmk999
     
  7. Feb 13, 2013 #6

    SteamKing

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    The Io values are the second moments (inertia) values of each part about its own center of gravity. For a rectangle, Io = (base/12)*(height^3)

    In the calculations of Ina, the parallel axis theorem says to subtract the total area A times the square of the distance to the NA. Since the distance to the NA = Mom / Area, then we can rewrite A*d^2 = Mom^2 / A. In this fashion, when dealing with large numbers, some precision is maintained which would be lost by calculating Ad^2.
     
  8. Feb 13, 2013 #7
    I understand your explanation

    My calculation below shows the inertia calculated about the vertical axis, is this correct?

    Item Area x-bar A*X A*Xsq Ixx

    part A 1500 75 112500 8437500 2812500

    part B 2920 5 14600 73000 24333.3

    part C 1000 50 50000 2500000 833333

    5420 32.67 177100 11010500 3670166


    IYY 3670166+11010500-177100sq/5420

    IYY = 8893875mm4

    Thanks for your help

    I now want to calculate radius of gyration about xx and yy. Is there a similar format and would you be happy to check my calc's once I've done them.

    hmk999
     
  9. Feb 13, 2013 #8

    SteamKing

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    The radius of gyration is a simple calculation:

    rxx = sqrt (Ixx/A)

    ryy = sqrt (Iyy/A)

    You already have calculated the values Ixx and Iyy and A required.
     
  10. Feb 13, 2013 #9

    SteamKing

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    Your Iyy calculation looks good.
     
  11. Feb 14, 2013 #10
    https://skydrive.live.com/redir?resid=22F7BBABD7DEE83B!138&authkey=!AEL3SAU9F1BCW-Y

    See attached link with updated calculations

    Hi can you give me a little more advice on this.

    Looking at the information I now have for the C section beam. I am trying to understand where the maximum stress points would be under load, I assume it would be the outer faces of the top and bottom flanges of the C section?

    If the beam was put under different uniform loads of say 8 tons and 12 tons. What is the formula to work out the max stress for this cross sectional area? I am trying to work out and understand what minimum size the section can be under various loads. Increasing the height and material thickness will change the value of I inertia. But how does this relate to working out max loads you can apply?

    I also notice we have Y = distance from the neutral axis which comes into the equation. I assume it’s the distance from the NA to the outer surfaces of the section, also where I think the max stress would occur.

    Can you please help me?

    The section is mild steel.

    Secondly, I understand you would have to calculate across the beam (simply supported beam macauleys method for the max stress and deflection, bending moments etc. Just trying to understand if you can calculate from the cross section only, when under load, what is the minimum size the C section can be?

    Thanks for your help
    Hmk999
     
  12. Feb 14, 2013 #11

    SteamKing

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    I have one small comment to make about your inertia calculations.

    The gyradius is not the same as the radius of curvature. The gyradius is a property of the cross section alone, while the radius of curvature deals with how a beam would deflect under the action of a bending moment.

    There are some calculations you can make about the maximum stress using only the cross section. You must make some assumptions about the axial forces, shearing forces, and any bending or twisting moments applied to the beam.

    If you wish to analyze a beam constructed from a C section under various loadings, I would be happy to review any calculations you make.

    I would also like to know if you are studying strength of materials, and if this work is related to that study.
     
  13. Feb 17, 2013 #12
    https://skydrive.live.com/redir?resid=3B1281F72DB8729B!117&authkey=!AFfQ3vG7MAQH8X8

    Hi, if you view the above link, I have created a diagram of half the beam as the beam is symmetrical, to the left of the diagram you will see the centreline and I have also placed a bending moment on the centreline.

    Can you please advise if the equation is correct for the bending moment? If not can you advise where I'm going wrong?

    I also want to create a equation for the deflection but trying to sort bending moment first.

    I've been struggling to get my head round this macauleys method so any help much appreciated.

    Not studying anything at the moment, I have an opportunity to learn some FEA at work and just trying to get a head start on some beam theory.

    Thanks
     
  14. Feb 18, 2013 #13

    SteamKing

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    Bending moments are usually of interest when they are applied along the length of the beam rather than across the breadth of the cross section.
     
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