支点反力Rをそのままにして用いる.
\[\begin{array}{}
(0\leqq x \leqq a) \hspace{20px} M=Rx
\\
(a\leqq x \leqq l) \hspace{20px} M=Rx-P(x-a)
\end{array}\]
曲げによるひずみエネルギー$U$を$R$で微分すれば,$R$方向のたわみが求まる.
このたわみが$0$(ゼロ)なので,
\[
\begin{align}
\frac{\partial U}{\partial R}&=\frac{\partial}{\partial R}\left\{\int_0^l\frac{M^2}{2EI}dx\right\}\\
&=\frac{1}{2EI}\int_0^l\frac{\partial}{\partial M}M^2\frac{\partial M}{\partial R}dx\\
&=\frac{1}{EI}\int_0^lM\frac{\partial M}{\partial R}dx\\
&=\frac{1}{EI}\left\{\int_0^a(Rx)xdx+\int_a^l\left(Rx-P(x-a)\right)xdx\right\}\\
&=0
\end{align}\]
を解いて,
\[\begin{align}
R&=\frac{2l^3-3al^2+a^3}{2l^3}P\\
&=\frac{(3l-b)b^2}{2l^3}P
\end{align}\]