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Deflections of a bar

  1. Jan 28, 2013 #1

    I am looking over deflection in materials and came across the following for a bar u(x,t) and I need some help in understanding it:

    F(x,t) = ρA[itex]\frac{∂^{2}u}{∂t^{2}}[/itex] + EI [itex]\frac{∂^{4}u}{∂x^{4}}[/itex]

    where ρ is the density of the bar, A is the cross-section, F is the force per unit length, E is Youngs modulus and I is the moment of inertia for the cross-section of the bar.

    What I can't understand in this is the [itex]\frac{∂^{2}u}{∂t^{2}}[/itex] and [itex]\frac{∂^{4}u}{∂x^{4}}[/itex].

    What do they mean and how do you find out what u(x,t) is to work out the 2nd and 4th diritive of it?
  2. jcsd
  3. Jan 28, 2013 #2


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    From the equations, it appears that the second part of the RHS is the normal Euler-Bernoulli beam equation, where u is the deflection of the beam as a function of position along the length of the beam. In total, this equation appears to be the dynamic form of beam response.

    As for u(x,t), that will depend on the loading on the beam.
  4. Jan 28, 2013 #3
    I agree with SteamKing.

    The basic equation of motion for a vibrating bar of uniform cross section is

    [tex]EI\frac{{{\partial ^4}u}}{{\partial {x^4}}} = - \mu \frac{{{\partial ^2}u}}{{\partial {t^2}}}[/tex]

    Where μ is the mass of bar per unit length.

    This has a general solution

    u(x) = Aeax + Be-ax + Csinax + Dcosax


    [tex]a = \sqrt[4]{{\frac{{\mu {\omega ^2}}}{{EI}}}}[/tex]

    The values of A, B, C & D are determined by the boundary (support) conditions.
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