Deriving the Bending Energy equation for Eulers Elastica In many (all?) papers regarding elastic curves the bending energy for the elastica is given by $$B[\gamma] = \int_{\gamma} \kappa^2(s)ds$$ where $\gamma$ denotes a planar curve of fixed length and clamped endpoints. Here $s$ denotes the arc length of the curve and $\kappa$ is the (signed) curvature. 
I want to know where this equation comes from and how it is derived. Here is one of the papers I am reading: https://arxiv.org/pdf/1710.05890.pdf In all of the material I have found the above bending energy is given as a definition and no motivation for it is given. Any help is greatly appreciated.
 A: Of course, you are missing some physical constants in your formula: you should say that the 
elastic energy is proportional to the integral of the squared curvature.
To derive that, take a segment of a bent thin beam  whose barycentric axis describes
the given curve.
Take an infinitesimal sector $ds=Rd\alpha$ of the beam, with curvature radius $=R$, and compute the strain of a small fiber of the beam
at distance $x$ from the barycentric axis: you will find that is $\epsilon=((R+x)d\alpha-Rd\alpha)/(Rd\alpha)=x/R)$.
The corresponding elastic energy per volume will be $1/2E*(x/R)^2$, which integrated all over the cross section will give $1/2EI_y/R^2ds$.
Brief explanation
Since you asked for that I am trying to give some hint on how to start and understand the subject.
Then have a look at this link and at this other one for a better assessment.

If you take a thin beam and imagine to cut a piece out of it at $s$ and $s+ds$, the coehesion forces acting on each small area
of the cross section ,i.e. the stress $\sigma$ (thus actually a pressure distribution), will translate into a resultant normal force $N$ (tension), 
tangential force $T$ (shear) and a moment of them $M$ (bending moment). We omit here the other possible resultant which is torque.
These resultants, taken on both sides shall equilibrate with the external forces (load $ Fds$ and the reactions at the support points).
The "slender beams" approximation (that is for beams relatively thin compared to the length) says that , when laterally loaded, 
the deformation is mainly due to the moment: we can approximate the cohesion forces with a triangular distribution in the plane of
bending, with negligible resulting tension and shear.  The strain $\epsilon=\Delta l /l$ of each elementary axial fiber will have the same triangular
distribution. The deformation of the $ds$ sector will result in a circular "wedge" with radius $R$ corresponding to the curvature.   
The strain $\epsilon$ as told equals $x/R$, the stress $\sigma$ is linearly related to it by the Young's modulus, the elastic energy per volume
of each elementary fiber will be $1/2\sigma \epsilon$, the total is thus $1/2\sigma \epsilon dl dA=1/2E x^2/R^2 dl(x) dA(x)$.
Summing over the cross section will give $1/2EI_y/R^2ds$.
