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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Virtual work in finite plane bending of Euler-Bernoulli beam

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