$(1)$ \[ U=\int_0^l\frac{M^2}{2EI}dx \] $(2)$ \[\begin{align} &\textrm{AC}間(0\leqq x\leqq\frac{l}{3})\hspace{20pt}M=-Px\\ &\textrm{CB}間(\frac{l}{3}\leqq x\leqq l)\hspace{20pt}M=-Px+R_c\left(x-\frac{l}{3}\right) \end{align}\] $(3)$ \[ \delta=\frac{\partial U}{\partial P}\] ひずみエネルギー$U$を外力$P$で偏微分すると，外力$P$の負荷されている点の負荷方向の変位が求まる． $(4)$ $\delta_c=0$を解けば良い \[\begin{align} \delta_c&=\frac{\partial U}{\partial R}\\ &=\frac{1}{EI}\int_0^lM\frac{\partial M}{\partial R}dx\\ &=\frac{1}{EI}\left\{\int_0^{\frac{l}{3}}(-Px)\times0\times dx+\int_{\frac{l}{3}}^l\left(-P_x+R_c\left(x-\frac{l} {3}\right)\right)\left(x-\frac{l}{3}\right)dx\right\}\\ &=\frac{1}{EI}\left[-\frac{P}{3}x^3+\frac{Pl}{6}x^2+R_c\left(\frac{1}{3}x^3-\frac{1}{3}lx^2+\frac{l^2}{9}x\right)\right]_{\frac{l}{3}}^l\\ &=\frac{l^3}{EI}\left(\frac{-756}{4374}P+R_c\frac{8}{81}\right)\\ &=0 \end{align}\] より\[ R_c=\frac{756}{4374}P\times\frac{81}{8}=\frac{7}{4}P \] $(5)$ \[ \delta_A=\frac{\partial U}{\partial P}=0\\ \delta_C=\frac{\partial U}{\partial R_c}\hspace{20px}\\ を解く\]