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Virtual work in finite plane bending of Euler-Bernoulli beam

  1. Jul 17, 2015 #1
    Please refer to the following image, which shows a portion of the deformed centerline of a beam in its equilibrium configuration with a uniformly distributed load.

    The stress resultants are the axial forces T, transverse shears Q, and bending moments M at sections 1 and 2, with the rotations being relative to the horizontal.

    We apply infinitesimal virtual displacements from the equilibrium configuration, with normal and tangential components [itex]\delta u_n[/itex] and [itex]\delta u_t[/itex] as functinos of arc length s.

    If the beam were straight, then the incremental rotation is:

    [itex]\delta \theta = \frac{d\delta u_n}{ds}[/itex]

    And the incremental strain would be:

    [itex]\delta \epsilon = \frac{d\delta u_t}{ds}[/itex]

    I understand that part.

    Now, given that the beam is not straight in its equilibrium position:

    [itex]\delta \theta = \frac{d\delta u_n}{ds} + \frac{d\theta}{ds}\delta u_t[/itex]

    [itex]\delta \epsilon = \frac{d\delta u_t}{ds} - \frac{d\theta}{ds}\delta u_n[/itex]

    It's not intuitive for me how curvature multiplied by displacement gives the extra terms. How are they obtained?
  2. jcsd
  3. Jul 22, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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