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Establishing Neutral Axis & Second Moment on a T bar beam

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm having difficluty in finding the second moment of area and neutral axis of this beam can anyone help

    2. Relevant equations



    3. The attempt at a solution

    I know that you have to divide the beam into two rectangles, I have used the formula bd3/12 and Izz, i did Izz = (1/12)6.4(38.1)^3 and the answer came to 29496.72

    surely this cant be right I am going wrong somewhere, can anyone help, it hasnt got to be in for a few weeks but I want to do this now.

    Can anyone help, I have attached the dimensions of the beam
     

    Attached Files:

  2. jcsd
  3. Nov 27, 2008 #2
    I have seen the question similar to mine by struggling but that does not explain it.

    Can anyone please help?
     
  4. Nov 27, 2008 #3
    You may want to describe the section of the beam, or post it on a photo album site, because the document appears to be waiting for approval.
     
  5. Nov 28, 2008 #4
    Sorry about that, please follow the link, any help would be appreciated.

    [​IMG]
     
  6. Nov 28, 2008 #5
  7. Nov 28, 2008 #6
    I understand all of that but what figure is d0 it says it is the distance of the central axis but if i multiply a larger number comes up.
     
  8. Nov 28, 2008 #7
    It would be much easier if you could post what you have attempted to do, so I can see how you arrived at your answer, or the problem of the 'large number'.
     
  9. Nov 28, 2008 #8
    Thanks for getting back to me,

    The figures I'm getting are 38.1 X 6.4 (243.84) + 31.7*6.4 (202.88)which is 446.72 this is fine as it is adding the two rectangles up however if I multiply like what Kazza said by 15.1 it gives me 6745.472, I am no where near the figure of 11.85 he got.

    The other formula I have is

    A X d + A2 X d2 divide by Area 1 + Area 2

    dont know which one I should use but i'm not getting 11.85 which seems to be the correct answer as it seems logical it may be right but I dont feel comfortable in just copying what his wrote because I dont know where he has got the figures from.

    Thanks
     
  10. Nov 28, 2008 #9
    The 11.85 he got was measured from the bottom edge of the flange (horizontal rectangle).

    All you need to do is to split the shape into two rectangles, namely the flange of 6.4*38.1 and the web of 6.4*31.7, as you mentioned. The 446.72 you obtained is perfect for the total area.

    Now you need to take moments of each area about the reference line, i.e. bottom edge of the flange, i.e. calculate [tex]\Sigma[/tex]AiDi.
    You have already calculated Ai of the individual rectangles.
    Now you need the distance of the centroid of each rectangle from the reference line.

    Divide this sum by 446.72 (sum of areas) should give you 11.85 as you expected.
     
  11. Nov 28, 2008 #10
    Sorry but I still cant do it, what do you eactly mean by moments of the refrence line,

    is it 243.84 * 3.2 + (202.88 * 15.85) + 6.4 = 4002.336

    if I divide that by 446.72 it gives 8.95 still not close enough what am I doing wrong.
     
  12. Nov 28, 2008 #11
    By the way the 15.85 is my refrence line which is the centroid
     
  13. Nov 29, 2008 #12
    any body?
     
  14. Nov 29, 2008 #13
    Thanks mathmate, your guidance has enabled me to crack the question I now have the answer to 11.85 and do undersatnd it now.

    Just another question how do you calculate the strain at the neutral axis, do i need to use another formula for this?
     
  15. Nov 29, 2008 #14
    Yes, you'd need another formula to calculate the stress, [tex]\sigma[/tex], at different places in the section.
    Stress is defined as force divided by the area, and strain,[tex]\epsilon[/tex], is the unit deformation, i.e. deformation divided by the length.
    In the absence of axial forces, the stress at the neutral axis is zero, therefore the strain also.
    The stress,[tex]\sigma[/tex], is given by the formula:
    [tex]\sigma[/tex]=Mc/I, where
    M=bending moment,
    I=moment of inertia of cross section, and
    c=distance from neutral axis.

    This means that the further away from the neutral axis, the greater is the stress. This formula assumes plane stress, meaning that the deformations across the section remain in a plane.

    The strain at different points is simply
    [tex]\epsilon[/tex] = [tex]\sigma[/tex]/E, where E is the modulus of elasticity.

    You said you have obtained the neutral axis. Did you calculate the moment of inertia of the section? What did you get?

    Just a side question: do you have any textbooks on Strength of Materials? If not, I can recommend a few that you can certainly get from the library:

    Strength of Materials, by Stephen Timoshenko (the classic, but rather expensive)
    http://www.amazon.com/Strength-Materials-Part-2/dp/0898746213

    Strength of Materials, by J. P. Den Hartog, an excellent work, available in Dover edition
    http://www.amazon.com/Strength-Materials-J-Den-Hartog/dp/0486607550

    Strength of Materials, By Pytel and Singer, a standard textbook, very affordable.
    http://www.amazon.com/Strength-Materials-Andrew-Pytel/dp/0060453133
     
  16. Nov 29, 2008 #15
    Thanks for the info about the books I have checked the uni library website and they are available to borrow so will do that monday.

    Did you mean second moment of area?
     
  17. Nov 29, 2008 #16
    You're right, it's second moment of area that I meant.

    Sorry I fell into that bad habit, as quoted in Wiki
    http://en.wikipedia.org/wiki/Second_moment_of_area
    It continued to say:
    which made me feel better!
     
  18. Nov 29, 2008 #17
    Second Moment of Area using parallel axis theorem

    I = Bd3
    12

    Rectangle A

    Y = 6.4 + 38.1 -11.85
    2
    Y= 13.6

    Bd3 + (13.6)2
    12

    = 1017.2mm4

    Rectangle B

    Y = 6.4 + 31.7 -11.85
    2
    Y= 10.85

    Bd3 + (10.85)2
    12

    = 810.2mm4

    Total Second moment of area = 1827mm

    is this correct?
     
  19. Nov 29, 2008 #18
    The second moment of area of a rectangle is
    I=bd3/12
    where b is the dimension parallel to the axis passing through the centroid, and
    d is the dimension perpendicular to the centroid.
    In the example given, rectangle A has a dimension of 38.1 and height of 6.4. The axis is horizontal passing through the centre of the rectangle at 3.2 from the bottom edge.
    So
    b=38.1 (parallel to horizontal axis)
    d=6.4 (perpendicular to axis)
    and
    Ia=38.1*6.43/12=832.3072
    Ib=second moment of inertia of rectangle (31.7*6.4) B
    Let
    Aa=area of rectangle A = 38.1*6.4=243.84
    Ya=distance of centroid of rectangle A from bottom flange= (6.4/2)=3.2
    Ab=area of rectangle B
    Yb=distance of centroid of rectangle B from bottom flange
    The distance of centroidal axis from the bottom flange (bottom of rectangle A)
    y=(Aa*Ya+Ab*Yb)/(Aa+Ab)=11.85 (as calculated before)

    The parallel axis theorem states that the second moment of area about an axis
    parallel and at a distance d from the centroidal axis is
    I'=Ic+A*d2
    Ic=second moment of inertia about its centroidal axis
    I'=second moment of inertia about the new axis.

    So the Total second moment of area is
    I=Ia+Aa*(Ya-y)2+Ib+Ab*(Yb-y)2
    =5#00#.865

    # represents a missing digit.

    There should be enough information for you to proceed with your calculations. In addition, here is a link that shows you how to tabulate your calculations, and hence minimizing errors.
    http://www.cen.bris.ac.uk/personal/cenaa/Nick's Web site/notes/Mechanics I/tutorial.pdf
     
  20. Nov 29, 2008 #19
    Thank you very much for that, I'm starting to learn it now may be you should consider being a lecturer, you have done more explaining than my own lecturers.

    I will have a go at that tonight, by the way the link you pasted does not work it says its a unkown file.

    Would it be possible to repaste it, thanks once again, you have been very kind.
     
  21. Nov 29, 2008 #20
    Thank you for the appreciation. It's a pleasure to know that my help is of some use.

    Perhaps the site was down at the moment you tried it. I tried it a moment ago and it worked.
    Also, if you copy and paste, it may not work, because it will be missing a few details. You just have to double click on the link. If you are worried about double-clicks, you can hover over the link, and copy by hand on the bottom left, of the window, which gives you the complete address without double-clicking. (that works for firefox, probably IE as well).

    Here it is again, just in case.
    http://www.cen.bris.ac.uk/personal/cenaa/Nick's Web site/notes/Mechanics I/tutorial.pdf
     
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