Castigliano: beam, redundant moment at A

Reading of the figure: A fixed; B is 2 m from A where a couple of 10 kNm (taken clockwise) is applied; C is a roller 2 m from B; CD is a 1 m overhang with 6 kN/m. EI is constant.

The fixity at A gives one redundant. Release the rotation at A and take the fixed-end moment $M_A$ (clockwise on the beam) as the redundant, so the beam becomes simply supported at A and C with an overhang. The condition is that the slope at A is zero:

$$\frac{\partial U}{\partial M_A} = \theta_A = 0$$

            10 kNm (cw)        6 kN/m
 A ||--------B--------C vvvvvvvv D
 (fixed)              ^ (roller)
 |<-- 2 -->|<-- 2 -->|<-- 1 -->|

Reaction at A in terms of $M_A$

Moments about C (the overhang load is 6 kN at 0.5 m from C):

$$4R_A + M_A + 10 + 6(0.5) = 0 \ \Rightarrow\ R_A = -\frac{M_A + 13}{4}$$

$\dfrac{\partial R_A}{\partial M_A} = -\dfrac{1}{4}$ ($R_A$ is positive upward).

Bending moments (x from A, sagging positive)

$$\begin{aligned} 0 \le x \le 2:\quad M &= M_A + R_A x, & \frac{\partial M}{\partial M_A} &= 1 - \frac{x}{4} \\ 2 \le x \le 4:\quad M &= M_A + R_A x + 10, & \frac{\partial M}{\partial M_A} &= 1 - \frac{x}{4} \end{aligned}$$

The overhang CD does not depend on $M_A$ (its moment comes from the UDL alone), so it adds nothing to the integral.

Castigliano's theorem

$$\int_0^4 M\left(1-\frac{x}{4}\right)dx = 0$$

Evaluating each term with $M = M_A - \dfrac{(M_A+13)x}{4} + 10\,[x>2]$:

$$\begin{aligned} M_A\int_0^4\left(1-\tfrac{x}{4}\right)^2dx &= \tfrac{4}{3}M_A \\ -\tfrac{13}{4}\int_0^4 x\left(1-\tfrac{x}{4}\right)dx &= -\tfrac{13}{4}\cdot\tfrac{8}{3} = -\tfrac{26}{3} \\ 10\int_2^4 \left(1-\tfrac{x}{4}\right)dx &= 10(0.5) = 5 \end{aligned}$$ $$\tfrac{4}{3}M_A - \tfrac{26}{3} + 5 = 0 \ \Rightarrow\ M_A = \tfrac{11}{4} = 2.75\ \text{kNm}$$

Other reactions

$$R_A = -\frac{2.75+13}{4} = -3.9375\ \text{kN}\ (\text{downward}), \qquad R_C = 6 - (-3.9375) = 9.9375\ \text{kN}\ (\uparrow)$$

Bending moments (kNm)

Answer: $M_A = 2.75$ kNm (clockwise on the beam, provided by the wall), $R_A = 3.938$ kN downward, $R_C = 9.938$ kN upward.