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Equivalent stiffness of a beam

  1. Sep 22, 2011 #1
    Hi all,

    I am trying to understand the concept of equivalent stiffness of a beam. As I see it, the equivalent stiffness is the stiffness of a linear spring that would deflect the same amount under the same load. For a cantilever beam with a load P and a deflection [itex]\delta[/itex] at the free end, if we just look at the deflection of the free end, and somehow lump the stiffness and elastic properties of the beam into the term [itex]k_{eq}[/itex], it's motion will be the same as a linear spring with stiffness [itex]k_{eq}[/itex] when the same load P is applied to the spring.

    My vibrations textbook mentions the [itex]k_{eq}[/itex] for a cantilever with a moment applied to the free end as [itex]\frac{EI}{L}[/itex]. Assuming that the spring being considered is a linear torsional spring, how do we interpret this? I am thinking it goes something like - 'The equivalent stiffness of a cantilever beam with a moment at the free end is the stiffness of a linear torsional spring that would coil by an angle say [itex]\theta[/itex] when the same moment is applied to it.' Now, is [itex]\theta[/itex] the same as the tip deflection [itex]\delta[/itex] or is it the local slope at the free end ie [itex]\theta \approx \tan(\theta) = \frac{dy}{dx}[/itex]

    Thanks a lot for the help!

    PS - Any suggestions for books that explain the equivalent stiffness concept well?
     
  2. jcsd
  3. Sep 22, 2011 #2

    AlephZero

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    [itex]\theta[/itex] is the local slope at the free end.

    Of course there is also a displacement at the free end, which you can find from the fact that the curvature of the beam is constant. The geometrical relationship between the slope and the displacement at the free end will depend on the length of the beam, but not on E or I.
     
  4. Sep 28, 2011 #3
    Sorry for the late response.

    Thanks for clarifying that. Also, does that mean that if we know the moment acting at a location x, we can calculate the slope of the curved beam at that point due to the moment?
     
  5. Oct 1, 2011 #4

    nvn

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    boeing_737: Regarding post 3, yes, it does. Regarding post 1, you have a cantilever having an applied tip moment, M. As AlephZero mentioned, the cantilever tip rotational stiffness is equivalent to a linear torsional spring at the cantilever tip having spring constant ktheta = E*I/L. The cantilever tip slope is theta = M/ktheta.

    Also, the cantilever tip transverse deflection stiffness is equivalent to a linear translational spring at the cantilever tip having spring constant k = 2*E*I/L^2. The cantilever tip transverse deflection is delta = M/k.
     
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