Castigliano: deflection at B of propped cantilever

Beam AC of span $L$: A fixed, C a roller, point load $P$ at B (mid-span). EI is constant. The beam is indeterminate to the 1st degree. First find the redundant $R_C$ and then the deflection at B.

          P
          v
 ||=======B=======o
 A (fixed)        C (roller, R_C)
 |<-- L/2 -->|<-- L/2 -->|

Step 1: Bending moments (x measured from C)

$$\begin{aligned} CB\ (0 \le x \le L/2):\quad & M = R_C\,x \\ BA\ (L/2 \le x \le L):\quad & M = R_C\,x - P\left(x - \frac{L}{2}\right) \end{aligned}$$

Step 2: Redundant reaction

The deflection at C is zero, so $\dfrac{\partial U}{\partial R_C} = 0$:

$$\int_0^{L/2} R_C\,x^2\,dx + \int_{L/2}^{L}\left[R_C\,x - P\left(x-\frac L2\right)\right]x\,dx = 0$$ $$R_C\frac{L^3}{3} - P\left[\frac{x^3}{3} - \frac{Lx^2}{4}\right]_{L/2}^{L} = 0$$ $$R_C\frac{L^3}{3} - P\,\frac{5L^3}{48} = 0 \ \Rightarrow\ R_C = \frac{5P}{16}$$

(The bracket equals $\left(\dfrac{L^3}{3} - \dfrac{L^3}{4}\right) - \left(\dfrac{L^3}{24} - \dfrac{L^3}{16}\right) = \dfrac{L^3}{12} + \dfrac{L^3}{48} = \dfrac{5L^3}{48}$.)

Step 3: Deflection at B by Castigliano's theorem

The load $P$ acts at B, so

$$\delta_B = \frac{\partial U}{\partial P} = \frac{1}{EI}\int M\,\frac{\partial M}{\partial P}\,dx$$

Because $\partial U/\partial R_C = 0$, the dependence of $R_C$ on $P$ does not add anything, so only the explicit $P$ in $M$ is differentiated: $\dfrac{\partial M}{\partial P} = -\left(x - \dfrac L2\right)$ for $L/2 \le x \le L$ and $0$ for the part CB.

With $R_C = 5P/16$:

$$\delta_B = \frac{1}{EI}\int_{L/2}^{L}\left[\frac{5P}{16}x - P\left(x - \frac L2\right)\right]\left[-\left(x-\frac L2\right)\right]dx$$

Let $u = x - L/2$ (from 0 to $L/2$), so $x = u + L/2$:

$$\delta_B = \frac{P}{EI}\int_0^{L/2}\left[u - \frac{5}{16}\left(u + \frac L2\right)\right]u\,du = \frac{P}{EI}\int_0^{L/2}\left[\frac{11}{16}u^2 - \frac{5L}{32}u\right]du$$ $$\delta_B = \frac{P}{EI}\left[\frac{11}{48}\cdot\frac{L^3}{8} - \frac{5L}{64}\cdot\frac{L^2}{4}\right] = \frac{P L^3}{EI}\left(\frac{11}{384} - \frac{5}{256}\right)$$ $$\delta_B = \frac{PL^3}{EI}\cdot\frac{22 - 15}{768} = \frac{7PL^3}{768\,EI}$$

Answer: the deflection at B is $\delta_B = \dfrac{7PL^3}{768\,EI}$ downward; $R_C = \dfrac{5P}{16}$.