Castigliano's second theorem: slope at B

Castigliano's second theorem: $\theta=\partial U/\partial M$. The beam is a propped cantilever (A fixed, B on a roller/hinge support) with a clockwise couple $M$ at B. The vertical reaction $R_B$ is the redundant, found from $\partial U/\partial R_B=0$ because the support does not settle.

Take $x$ measured from B towards A (B at $x=0$, A at $x=l$), $R_B$ upward.

Bending moment (sagging +)

$$M_x = M + R_Bx,\qquad \frac{\partial M_x}{\partial R_B}=x,\qquad \frac{\partial M_x}{\partial M}=1$$

Step 1: find $R_B$

$$\frac{\partial U}{\partial R_B}=\frac{1}{EI}\int_0^l(M+R_Bx)\,x\,dx=0 \;\Rightarrow\; \frac{Ml^2}{2}+\frac{R_Bl^3}{3}=0 \;\Rightarrow\; R_B=-\frac{3M}{2l}$$

(i.e. $3M/2l$ acting downward.)

Step 2: slope at B

$$\theta_B=\frac{\partial U}{\partial M}=\frac{1}{EI}\int_0^l(M+R_Bx)(1)\,dx=\frac{1}{EI}\left[Ml+R_B\frac{l^2}{2}\right]$$ $$\theta_B=\frac{1}{EI}\left[Ml-\frac{3M}{2l}\cdot\frac{l^2}{2}\right]=\frac{1}{EI}\left[Ml-\frac{3Ml}{4}\right]=\frac{Ml}{4EI}$$

Answer: $\theta_B=\dfrac{Ml}{4EI}$, in the same (clockwise) sense as the applied moment. The fixed far end stiffens the beam, so the slope is smaller than the simply-supported value $Ml/3EI$.