Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivation of resonance fixed-fixed beam

  1. Sep 7, 2011 #1
    I'm trying to derive the resonance frequencies for a simple fixed-fixed beam, as opposed to a simply-supported beam.

    I'm working off the following references:
    1. http://emweb.unl.edu/Mechanics-Pages/Scott-Whitney/325hweb/Beams.htm
    But this is for a fixed-free cantilever beam.

    2. alrafidain.engineering-coll-mosul.com/files/132.pdf

    But this is for a simply-supported (or 'pinned') beam.

    Both sources follow the same derivation steps.
    If we write the differntial equation as follows:
    [tex] EI\frac{\partial^{4}z(x,t)}{\partial x^{4}} = \rho A \frac{\partial^{2}z(x,t)}{\partial t^{2}} [/tex]
    Than for my application I would like to state the following boundary conditions:
    [tex] 1, z(0,t) = 0 [/tex]
    [tex] 2, z'(0,t) = 0 [/tex]
    [tex] 3, z(L,t) = 0 [/tex]
    [tex] 4, z'(L,t) = 0 [/tex]
    For a fixed - fixed beam.

    However, I can't seem to derive this to a manageable equation.

    The general solution is in the form of:
    [tex] z(x,t) = (A \cos(\omega t)+ B \sin(\omega t)) \cdot ( C_1 \sin(\alpha x) + C_2 \cos(\alpha x) + C_3 sinh(\alpha x) + C_4 cosh(\alpha x)) [/tex]

    Assuming the time term is not zero.
    Boundary Condition 1 gives:
    [tex] C_2 + C_4 = 0 [/tex]
    BC 2:
    [tex] C_1 + C_3 = 0 [/tex]
    BC 3:
    [tex] C_1 sin(\alpha L) + C_2 cos(\alpha L) -C_1 sinh(\alpha L) - C_2 cosh(\alpha L) = 0 [/tex]
    BC 4:
    [tex] C_1 cos(\alpha L) - C_2 sin(\alpha L) -C_1 cosh(\alpha L) - C_2 sinh(\alpha L) = 0 [/tex]

    I can get an expression for C1 to C4 but it's incredibly long and I'm not sure what to do with it next. It doesn't derive to a nice equation as in the references.

    Could someone with expertise have a look and see if I'm stating my equations correctly. And maybe help me along with the derivations?
  2. jcsd
  3. Sep 7, 2011 #2
    Look again at your boundary conditions.

    I agree z(0) = z(L) = 0

    but why is

    z'(0) = z'(L) = 0
  4. Sep 7, 2011 #3


    User Avatar
    Science Advisor

    Standard conditions for a beam problem. The beam is clamped at the ends, so both the displacement and the slope must be zero there.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook