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In this article (already mentioned in this question) the dynamics of a planar elastic beam with "cantilever constrains" (one clamped end and one free end) is modeled.

Using the Euler-Bernoulli Beam theory the PDE describing the dynamics of the beam should be: $$ \gamma^2 \frac{\partial^4 y(s,t)}{\partial s^4} + \frac{\partial^2 y(s,t)}{\partial t^2}=0 $$ Where $y(s,t)$ in this formulation is the vertical displacement (expressed in a Cartesian reference) of the material point of the beam identified by the arch-lengh coordinate $s$ at time $t$. The distributed load along the beam it's assumed to be zero.

The boundary conditions should be expressed as:

  • At clamped end $s=0$:

$y(0,t)=0$

$\frac{\partial y(0,t)}{\partial s}=0 $

  • At free end $s=L$:

$\frac{\partial^2 y(L,t)}{\partial s^2}=0 $ That correspond to zero bending moment on the last cross section.

$\frac{\partial^3 y(L,t)}{\partial s^3}=0 $ That correspond to zero shear forces on the last cross section.

In the mentioned article a curvature formulation is used that hold in the hypothesis of small $y_{s}$, in this case one can write $k=y_{ss}$, where $k(s,t)$ identify the curvature of the beam. In this second formulation the previous PDE becomes:

$$ \gamma^2 \frac{\partial^4 k(s,t)}{\partial s^4} + \frac{\partial^2 k(s,t)}{\partial t^2}=0 $$

And its easier to treat in case of constant curvature initial conditions. The equivalent boundary conditions used are:

  • At free end $s=L$:

$\frac{\partial^2 y(L,t)}{\partial s^2}=0 \to k(L,t)0 $

$\frac{\partial^3 y(L,t)}{\partial s^3}=0 \to \frac{\partial k(L,t)}{\partial s}=0 $

And I'm fine with that.

  • At clamped end $s=0$:

$y(0,t)=0 \to \frac{\partial^2 k(0,t)}{\partial^2 s}=0 $ (?)

$\frac{\partial y(0,t)}{\partial s}=0 \to \frac{\partial^3 k(0,t)}{\partial^3 s}=0 $ (?)

Maybe I'm missing some trivial math step (I came from an engineering background) but these last transformation are not straight forward for me. Why is it possible to use such conditions?

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