Castigliano: reaction at C of L-frame

Reading of the figure: A (0, 0) is fixed at the bottom of the vertical member AB (length $L$), B is the corner, BC is the horizontal member (length $L$) carrying $W$ kN/m, and C is on a roller that gives a vertical reaction $R_C$. EI is constant. The frame has one redundant, $R_C$.

        W kN/m
  B ================= C  (roller)
  |                   o
  |
  | L
 A///   |<---- L ---->|

Bending moments (take $R_C$ upward)

• Member CB ($x$ from C): $M = R_C\,x - \dfrac{Wx^2}{2}$, $\ \dfrac{\partial M}{\partial R_C} = x$
• Member BA ($y$ from B downward): the moment is constant along the column, $M = R_C L - \dfrac{WL^2}{2}$, $\ \dfrac{\partial M}{\partial R_C} = L$

Castigliano's theorem

The roller does not move vertically, so $\dfrac{\partial U}{\partial R_C} = 0$:

$$\frac{1}{EI}\left[\int_0^L \left(R_C x - \frac{Wx^2}{2}\right)x\,dx + \int_0^L \left(R_C L - \frac{WL^2}{2}\right)L\,dy\right] = 0$$ $$\frac{R_C L^3}{3} - \frac{WL^4}{8} + R_C L^3 - \frac{WL^4}{2} = 0$$ $$\frac{4}{3}R_C L^3 = \frac{5}{8}WL^4 \ \Rightarrow\ R_C = \frac{15\,WL}{32}$$

Other reactions

$$V_A = WL - \frac{15WL}{32} = \frac{17\,WL}{32}\ (\uparrow),\qquad H_A = 0,\qquad M_A = \left|\frac{15WL^2}{32} - \frac{WL^2}{2}\right| = \frac{WL^2}{32}$$

The moment in the column is $-WL^2/32$, constant over its height, with tension on the outer face. The maximum sagging moment in the beam occurs at $x = 15L/32$ from C.

Answer: $R_C = \dfrac{15\,W L}{32}$ (upward).