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Beam bending strain field

  1. Sep 4, 2012 #1
    If I have a wide beam, parallel to the x axis, with its COM at the origin, then I want it to curve about the y axis, what would the elements of the strain tensor be?

    I have come to the conlusion that the beam would, for example,contract above x axis and expand below it. But I don't know how to describe strain relative to the y and z axes, and then how to translate this knowledge to the strain tensor.

    Thanks for your time.
     
  2. jcsd
  3. Sep 4, 2012 #2
    "wide beam" to me sounds like a shell.

    Reissner-Mindlin shell theory is what is used in Finite Element Analysis. They neglect the higher order terms but there is a lot of information on the subject.

    If you are more interested in analytical theory that is analogous to standard beam theory, then I believe "Kirchhoff-Love" would be what you are looking for.

    Hope that helps,
     
  4. Sep 5, 2012 #3
    Hello Lewis, you really need to supply more information.

    Wide beam? : What sort of wide beam?

    Wide flanged I beams are available for increased bending resistance and their properties are tabulated.

    There are code requirements and specifications for wide RC beams.

    I am going to assume a simple homogeneous rectangular wide beam of height h in the y direction and breadth b in the z direction.

    Yes you need also to specify the z direction - wide beams are 3 dimensional.

    A beam is considered wide when

    b >> h say 5 times or more.

    Under these conditions the material is not free to expand or contract in the lateral z direction under bending along the x direction about the y direction. (note I said direction not axis)

    In particular εz ≈ 0 at z=0 so


    [tex]{\varepsilon _z} = \frac{1}{E}[{\sigma _z} - \nu \left( {{\sigma _x} + {\sigma _y}} \right) \approx 0[/tex]

    Since h is small σy ≈ 0 So

    [tex]{\sigma _z} = \nu {\sigma _{{x_{z = 0}}}}[/tex]

    Thus

    [tex]{\varepsilon _x} = \frac{{1 - {\nu ^2}}}{E}{\sigma _x} = \frac{{1 - {\nu ^2}}}{{E{I_z}}}{M_z}y[/tex]

    In general the strain is reduced by a factor of [itex]{1 - {\nu ^2}}[/itex]

    Edit
    So you can see that the change is the insertion a modifying constant into your strain tensor, I will leave you to do this since you haven't provided any notation.
    You need to be careful here since my strain is engineering strain, not tensor strain, which is a factor of 1/2 different.

    You should also note that the sideways distribution may also depend upon the support conditions.
     
    Last edited: Sep 5, 2012
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