Castigliano: beam with yielding middle support

Beam ABC has $AB = BC = l$, hinged ends A and C, a spring-type support at B, and UDL $w = \dfrac{P}{2l}$ over $2l$ (total load $P$). Support B settles $\lambda R_B$, where $R_B$ is its upward reaction.

 w = P/2l
 vvvvvvvvvvvvvvvvvvvvvvvv
 o---------o---------o
 A         B(R_B)    C
 |<-- l -->|<-- l -->|

By symmetry $R_A = R_C = \dfrac{P - R_B}{2}$.

For $0 \le x \le l$ from A (the other half is the same):

$$M = \frac{P - R_B}{2}\,x - \frac{P}{4l}\,x^2, \qquad \frac{\partial M}{\partial R_B} = -\frac{x}{2}$$

Castigliano's theorem for the upward redundant $R_B$ gives the upward displacement of B. B moves down by $\lambda R_B$, so

$$\frac{\partial U}{\partial R_B} = -\lambda R_B$$ $$\frac{2}{EI}\int_0^l \left[\frac{P - R_B}{2}x - \frac{P}{4l}x^2\right]\left(-\frac{x}{2}\right)dx = -\lambda R_B$$

Evaluate the integral:

$$-\frac{1}{EI}\left[\frac{P - R_B}{2}\cdot\frac{l^3}{3} - \frac{P}{4l}\cdot\frac{l^4}{4}\right] = -\lambda R_B$$ $$\frac{(P - R_B)\,l^3}{6} - \frac{P l^3}{16} = EI\,\lambda R_B$$ $$\frac{5Pl^3}{48} - \frac{R_B l^3}{6} = EI\,\lambda\,R_B$$ $$R_B = \frac{5P}{8}\cdot\frac{1}{1 + \dfrac{6EI\lambda}{l^3}}$$

Answer: $R_B = \dfrac{5Pl^3}{8\,(l^3 + 6EI\lambda)}$ upward. For $\lambda = 0$ (rigid support), $R_B = \dfrac{5P}{8}$ (the continuous-beam value).