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Tapered beam deflection problem

  1. Dec 15, 2009 #1
    Hello everybody! So I have this beam beading problem I've been trying to figure out, (this is not a homework assignment) maybe someone can help. I have this beam that has a load at the midpoint. It's being simple supported on both ends and it's height tapers along it length on both sides. The base remains constant. I have experimental data on the max deflection at the midpoint, Force applied at midpoint and I'm looking for the experimental Young's Modulus (E) of the material. My first though was to apply d2y/dx2 =-M/EI => ymax =(-M/EI)dxdx. but I'm getting lost.. Can someone help?

    Thanks,
    Nauen
     
  2. jcsd
  3. Dec 17, 2009 #2

    nvn

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    nauen: You're almost there. Deflection y = (1/E)*integral[integral{[M(x)/I(x)]*dx}*dx], integrated from x = 0 to 0.5*L.
     
  4. Dec 18, 2009 #3
    Can I do it that way because this a tapered beam the bending is not going to be symmetrical? Will the equation still hold?
     
  5. Dec 19, 2009 #4

    nvn

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    Are the left and right sides of your beam symmetrical about a vertical axis passing through the beam midpoint?
     
  6. Dec 19, 2009 #5
    I attached a image showing the problem. So the beam is symmetrical about the y axis and tapered about the x axis. Because of this i would assume that the area with a smaller h would bend more. Since i have a force and a deflection at yc, would that change the way I construct my equation to find the E value.

    Thanks
     
  7. Dec 20, 2009 #6

    nvn

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    nauen: The midspan deflection would be as follows. I(x) means I is a function of x. In your case, solve for E.

    [tex]y_{\,\mathrm{midspan}} = \frac{1}{E}\left(-0.5{\cdot}L
    \int_{0}^{0.5\cdot L}\frac{M(x)}{I(x)}\,dx
    \;\;+\;\int_{0}^{0.5\cdot L}
    \int_{0}^{x}\frac{M(x)}{I(x)}\,dx\ dx\right)[/tex]
     
    Last edited: Dec 20, 2009
  8. Dec 21, 2009 #7
    Thanks nvn, But i can't figure out where the first integral came from.. when i worked it out my self, I only got a function of the second (double) integral. Could you explain it?


    THANK YOU, THANK YOU, THANK YOU
    -nauen
     
  9. Dec 21, 2009 #8

    nvn

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    It is an integration constant.
     
  10. Dec 21, 2009 #9
    ahhhhh....it is. Thank you again nvn
     
  11. Dec 22, 2009 #10
    Dear nauen and nvn,

    I am just reading your posts, but I am confused about what appears to be the conclusion about the deflection of a "tapered" beam being the same as the deflection of a simple beam of uniform thickness. A tapered beam(as I understand it) would be flat on the underside, and arc shaped on top, and would definately deflect less, would it not? Am I missing something? Are you guys just estimating? I am assuming the mathematics for a non-uniform beam would be trickier.

    Note that I have seen beam bridges that employ a thicker "web" towards the center than at the end points. I think the tapered beam problem is similar.

    By the way, I would also appreciate it if you guys can refer me to a book with high quality information on this topic of beams and such.


    Thank you.
     
  12. Dec 28, 2009 #11

    PhanthomJay

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    For a beam of uniform shape, I(x) would be constant; the deflection therfore would not be the same.
    Not necessarily arc shaped, but the tapered beam would deflect more than a uniform beam designed to carry the same max moment at midspan (the latter of which would be much stiffer).
    It is. For a uniform beam, I is constant. For a tapeerd beam, I is a function of x.
    The 'taller' section at midspan is used generally for stress rather than deflection purposes, placing the largest cross section at mid span wher the max load occurs, and tapering it down at the ends for economic and other purposes.
     
  13. Dec 25, 2010 #12
    I have an identical situation posed by nauen and responded to by nvn. It is a simply supported beam with one load in the center. The beam cross section varies with x but is symmetrical across the y axis. In one situation the beam width is constant and the height varies uniformly and in another situation the beam height is constant but the width varies uniformly. In both cases the thicker/wider sections are in the middle. I am looking for the deflection at discrete values of x and not just at the midpoint. The idea is to play with the beam profile and try to approximate the deflection curve to be as close to a circular arc as possible.

    I am not a mathematician so it will help if I get a solution after solving the integrals. The I(x) could be considered as (C+K*x)(h3)/12 in the case of varying width and uniform height and as (b*(C+K*x)3)/12 in the case of varying height and uniform width. C and K are constants and b is the uniform width of the beam and h is the uniform height of the beam as the case may be.


    If anyone can help it will be highly appreciated.
     
  14. Dec 26, 2010 #13
    Are you sure?
     
  15. Dec 26, 2010 #14
    Yes, I do understand that stiffer mid sections are used to withstand stress. However, in this case the "taller" and "wider" sections at the middle of the beam (both of which contribute to a higher moment of inertia) are intended to try and achieve a more circular pattern of deflection. The problem for which I am trying to get a solution inherantly requires deflection which must be circular in geometry. Stress is not a consideration. If I can get the equations given by nvn further simplified afer after performing the integration I will be able to use them to achieve the desired result through numerical means.
     
  16. Dec 26, 2010 #15

    nvn

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    vsrao: In your case, change ymidspan in post 6 to y(x), for 0 ≤ x ≤ 0.5*L. And change the first and third 0.5*L to x. If you cannot solve the integrals analytically, you will need to solve it numerically.
     
    Last edited by a moderator: Dec 27, 2010
  17. Dec 26, 2010 #16
    I understood the changes that you have referred to and I also understand that I should substitute "I" with the two different functions of X that I had included in my post # 12. My problem is that I am not a mathematician and do not know how to take this further till I get a simpler equation. I have no idea how to solve integrals numerically. If you can help more, great. Otherwise, thanks for the trouble.
     
  18. Dec 27, 2010 #17
    The thicker web does not mean a taller cross section. The web is the part that connects the two flanges in an I beam. They could use the thicker web for several reasons and with out knowing the specifics its hard to tell. You use the taller section because you give yourself a longer moment arm between the extreme fibers reducing the stress and increasing the overall capacity. To be real honest I'm not sure why you would want a thicker web at the midsection. Shear failures in concrete beams usually occur at the ends, and in a steel beam you would just add stiffeners to resist buckling. So its hard to tell anything with out knowing firstly what the material is.
     
  19. Dec 27, 2010 #18

    PhanthomJay

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    I think there is a confusion in the words 'thicker web'. I took it to mean taller (deeper) web since the question was on tapering.
     
  20. Dec 27, 2010 #19
    For a simply supported beam with a central point load, plus perhaps a uniformly distributed load as well, the maximum bending moment is in the centre and the maximum shear ocurs at each support.

    Consequently any beam design has to be deep enough to resist both these loads.
    For reinforced concrete this usually means a (much thicker root (support) section.
    For trusses the reverse can be true as alternative shear resistance can be provided.

    This thread discusses both situations.

    https://www.physicsforums.com/showthread.php?t=443267&highlight=quebec
     
  21. Dec 27, 2010 #20
    Thicker web means the web is wider. Deeper web means the section is deeper. In some cases, the web is the only dimension that changes. The flanges in most precast concrete girders are kept the same to facilitate precasting and spacers are added to increase the depth of the section. If you make the web wider, you have to make the flanges wider to remain economical and avoid building new forms every project ->$$$.
     
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