Isotropic thin-plate bending energy ($\mathcal{U}=\frac{D}{2}\int [(\Delta w{)}^2-2(1-\nu )\det D^{2}w]$)

For a homogeneous isotropic plate of thickness $h$, Young's modulus $E$ and Poisson's ratio $\nu$, the bending stiffness is $D=E{h}^3/[12(1-{\nu }^2)]$. Its quadratic bending energy isIt follows by integrating the strain energy density $\frac{1}{2}{\tau }_{ij}{e}_{ij}$ through the thickness: linear bending strains are proportional to $z$, and ${\int }_{-h/2}^{h/2}{z}^{2}dz=h^3/12$. Taking a first variation and integrating by parts twice gives $D{\Delta }^{2}w=0$ without applied transverse load. Clamped boundary variations have both zero value and zero normal derivative.

For a plate occupying $x>0$, the boundary part of the first variation of its isotropic thin-plate bending energy is, up to an overall nonzero sign,At a free edge the variation and its normal derivative can be prescribed independently. Their coefficients must therefore vanish, giving the two displayed natural boundary conditions: zero bending moment and zero effective transverse shear. The mixed-derivative coefficient includes the derivative along the edge of the twisting moment; keeping only the normal derivative of $w_{xx}+\nu w_{yy}$ would miss that contribution.