= Solution
The net upward <force> is the sum of the contact <point forces>, since the distributed <normal contact force> vanishes in every regime here. Before contact it is zero. During isolated contact it is $R_0=12B(\alpha L-2H)/L^3$. For an interval of contact it is $R_{-s}+R_s=12BH/d^3$, with $d=3H/\alpha$. Therefore
$$
\boxed{F_{\mathrm{up}}(\alpha)=
\begin{cases}
0,&0\leq\alpha\leq2H/L,\\
\dfrac{12B}{L^2}\left(\alpha-\dfrac{2H}{L}\right),
&2H/L\leq\alpha\leq3H/L,\\
\dfrac{4B\alpha^3}{9H^2},&\alpha\geq3H/L.
\end{cases}}
$$
The formulae agree at both transitions. At $\alpha=3H/L$ they give $12BH/L^3$, and their first <derivatives> also agree there. As a check on the sign and normalization, integrating the <Euler-Bernoulli beam equation> over the whole <elastic beam> gives $F_{\mathrm{up}}=B(w'''(L^-)-w'''(-L^+))$, which yields the same values.
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