We write the equations of static equilibrium and then proceed as in the Castigliano's method, except that the force with respect to which we take the derivative is one of the statically indeterminate reactions.

\begin{aligned} &\sum{X}=0 & R_2=0\\ &\sum{Y}=0 & R_1+R_3-5ql=0 (1)^*\\ &\sum{M_1}=0 & M_U-3l*R_3+5ql*\frac{5}{2}l=0 (2)^*\\ &\int^{L_1}\frac{1}{E*I}M_{g1}*\frac{\partial M_{g1}}{\partial H}+\int^{L_2}\frac{1}{E*I}M_{g2}*\frac{\partial M_{g2}}{\partial H}=0\\ &M_{g1}=-\frac{1}{2}*qx^2 & M_{g2}=-\frac{1}{2}qx^2+R_3*(x-2l)\\ &\frac{\partial M_{g1}}{\partial R_3}=0 & \frac{\partial M_{g2}}{\partial R_3}=x-2l\\ \end{aligned}

The partial derivative of the elastic energy of the entire system with respect to the statically indeterminate quantity is equal to zero.

\begin{aligned} &\int_{2l}^{5l}[-\frac{1}{2}qx^2+R_3(x-2l)]*(x-2l)dx=0\\ &\int_{2l}^{5l}-\frac{1}{2}qx^2(x-2l)dx + \int_{2l}^{5l}R_3(x-2l)dx=0\\ &-\frac{297}{8}ql^4+9*R_3l^3=0 & R_3=\frac{33}{8}ql\\ \end{aligned}

We add the remaining reactions

\begin{aligned} &(1)^* R_1=5ql-R_3=\frac{7}{8}ql & (2)^* M_U=3R_3l-\frac{25}{2}ql^2=-\frac{1}{8}ql^2\\ \end{aligned}