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Deflection of an Elastic Beam

  1. Feb 21, 2004 #1
    When an elastic beam AB supports a block of weight W at a given point B, the deflection [tex]y_s[/tex] (static deflection) is proportional to W. Show that if the same block is dropped form a height h onto the end B of a cantilever beam, the maximum deflection [tex]y_m = y_s(1 + (1+\frac{2h}{y})^\frac{1}{2}).[/tex] Neglect the weight of the beam and any energy dissipated in the impact.

    I have:

    [tex]y_s = kW[/tex]

    T1 = 0
    V1 = mgh
    T2 = 0 when deflection is at a max
    [tex]V2 = -mgy_m[/tex]

    I'm pretty sure that V2 should also include the potential energy stored in the beam, but I don't know how to express that. Would it be similar to a spring? V(beam) = 1/2ky^2 ? That's my guess, but there must be a logical way of proving it.

    I think I can do the rest once I find the expression for the potential energy inside the elastic beam.
  2. jcsd
  3. Feb 26, 2004 #2

    Tom Mattson

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    Last edited by a moderator: Apr 20, 2017
  4. Feb 26, 2004 #3


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    Yes, there is. You have identified a force law that should look familiar (if you put the k on the other side). How do you derive the potential energy from that force law (think about F dot dx).
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