= Isotropic thin-plate bending energy
{title2=$\mathcal U=\frac D2\int[(\Delta w)^2-2(1-\nu)\det D^2w]$}
For a homogeneous isotropic plate of thickness $h$, <Young's modulus> $E$ and <Poisson's ratio> $\nu$, the bending stiffness is $D=Eh^3/[12(1-\nu^2)]$. Its quadratic bending energy is
$$
\mathcal U[w]=\frac D2\int_\Omega
[w_{xx}^2+2\nu w_{xx}w_{yy}+w_{yy}^2+2(1-\nu)w_{xy}^2],dx,dy.
$$
It follows by integrating the <strain energy density> $\tfrac12\tau_{ij}e_{ij}$ 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 $D\Delta^2w=0$ without applied transverse load. Clamped boundary variations have both zero value and zero normal derivative.
Back to article page
