= Solution
Let the contact interval be $[-s,s]$, where $w=-H$. Its <slope> and <curvature> vanish. <Continuity> of the <displacement field>, <slope> and <bending moment> at $x=s$ gives
$$
w(s)=-H,\qquad w'(s)=0,\qquad w''(s)=0.
$$
On the free interval $s<x<L$, the cubic <polynomial> satisfying these three conditions is $w=-H+C(x-s)^3$. Put $d=L-s$. The end <boundary conditions> become
$$
Cd^3=H,\qquad3Cd^2=\alpha.
$$
Consequently
$$
\boxed{d=\frac{3H}{\alpha},\qquad
s=L-\frac{3H}{\alpha},\qquad
C=\frac{H}{d^3}.}
$$
This requires $\alpha\geq3H/L$, with $s=0$ at the transition. The full <displacement field> is
$$
\boxed{w(x)=
\begin{cases}
-H,&|x|\leq s,\\
-H+H\left(\dfrac{|x|-s}{L-s}\right)^3,&s\leq|x|\leq L.
\end{cases}}
$$
It satisfies nonpenetration. In the open contact interval the distributed <normal contact force> is $R=Bw''''=0$, since the obstacle is flat and $p=T=0$. The <normal contact force> is concentrated at the two endpoints. At $x=s$, $w'''$ jumps from zero to $6H/d^3$; at $x=-s$, it jumps from $-6H/d^3$ to zero. Both upward <point forces> are therefore
$$
\boxed{R_{-s}=R_s=\frac{6BH}{d^3}>0.}
$$
At $s=0$ they merge into the single midpoint <point force> of the preceding regime. These formulae belong to the small-<slope> model; interpreting arbitrarily large $\alpha$ as a finite physical rotation would exceed its assumptions.
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