Castigliano: central deflection of simply supported beam

Castigliano's second theorem: the deflection at a point equals $\partial U/\partial P$, where $P$ is a load at that point. There is no point load at mid-span, so apply a fictitious load $P$ downward at the centre and set $P=0$ after differentiating.

Reactions

$$R_A=R_B=\frac{wL}{2}+\frac{P}{2}$$

Bending moment (x from A; the beam is symmetrical, so work with half and double)

For $0\le x\le L/2$:

$$M=\left(\frac{wL}{2}+\frac{P}{2}\right)x-\frac{wx^2}{2},\qquad \frac{\partial M}{\partial P}=\frac{x}{2}$$

Deflection at the centre

$$\delta_C=\frac{\partial U}{\partial P}=2\int_0^{L/2}\frac{M}{EI}\frac{\partial M}{\partial P}dx\Big|_{P=0} =\frac{2}{EI}\int_0^{L/2}\left(\frac{wLx}{2}-\frac{wx^2}{2}\right)\frac{x}{2}\,dx$$ $$=\frac{w}{2EI}\int_0^{L/2}\left(Lx^2-x^3\right)dx=\frac{w}{2EI}\left[\frac{Lx^3}{3}-\frac{x^4}{4}\right]_0^{L/2} =\frac{w}{2EI}\left[\frac{L^4}{24}-\frac{L^4}{64}\right]$$ $$=\frac{w}{2EI}\cdot\frac{(8-3)L^4}{192}=\frac{5wL^4}{384EI}$$

Answer: maximum central deflection $\delta_C=\dfrac{5wL^4}{384\,EI}$ (downward).