Castigliano: fixed moment, propped cantilever

Propped cantilever AB: A fixed, B a roller, span 12 m, loads 100 kN at 4 m and 50 kN at 8 m from A. EI is constant. Take the roller reaction $R_B$ as the redundant. Measure $x$ from B towards A.

         50 kN      100 kN
           v          v
 B o-------+----------+---------|| A (fixed)
 |<- 4 ->|<--- 4 --->|<--- 4 --->|

(Positions from B: 50 kN at 4 m, 100 kN at 8 m.)

Bending moments

$$\begin{aligned} 0 \le x \le 4:\quad & M = R_Bx, & \frac{\partial M}{\partial R_B} &= x \\ 4 \le x \le 8:\quad & M = R_Bx - 50(x-4), & \frac{\partial M}{\partial R_B} &= x \\ 8 \le x \le 12:\quad & M = R_Bx - 50(x-4) - 100(x-8), & \frac{\partial M}{\partial R_B} &= x \end{aligned}$$

Castigliano's theorem ($\delta_B = 0$)

$$\int_0^{12}R_Bx^2\,dx - 50\int_4^{12}(x-4)x\,dx - 100\int_8^{12}(x-8)x\,dx = 0$$

Evaluate each term:

$$\begin{aligned} R_B\frac{12^3}{3} &= 576\,R_B \\ 50\int_4^{12}(x^2 - 4x)\,dx &= 50\left[\frac{x^3}{3} - 2x^2\right]_4^{12} = 50\,(576 - 288 - 21.333 + 32) = 14933.3 \\ 100\int_8^{12}(x^2 - 8x)\,dx &= 100\left[\frac{x^3}{3} - 4x^2\right]_8^{12} = 100\,(0 - (170.667 - 256)) = 8533.3 \end{aligned}$$ $$576R_B = 14933.3 + 8533.3 = 23466.7 \ \Rightarrow\ R_B = 40.741\ \text{kN}$$

Moment at the fixed support

$$M_A = R_B(12) - 50(8) - 100(4) = 40.741(12) - 800 = -311.11\ \text{kNm}$$

The fixed-end moment is 311.11 kNm (hogging). The vertical reaction is $R_A = 150 - 40.741 = 109.259$ kN.

Answer: $M_A = 311.11$ kNm (hogging); $R_B = 40.741$ kN.