Castigliano: prop reaction with BMD

Propped cantilever AB: A fixed, B a roller, span 10 m, 50 kN at 4 m from A (6 m from B). EI is constant. Take $R_B$ as the redundant and measure $x$ from B.

              50 kN
                v
 B o-----------+---------|| A (fixed)
 |<--- 6 m --->|<- 4 m ->|

Bending moments

$$\begin{aligned} 0 \le x \le 6:\quad & M = R_Bx, & \frac{\partial M}{\partial R_B} &= x \\ 6 \le x \le 10:\quad & M = R_Bx - 50(x-6), & \frac{\partial M}{\partial R_B} &= x \end{aligned}$$

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

$$\int_0^{10}R_Bx^2\,dx - 50\int_6^{10}(x-6)\,x\,dx = 0$$ $$\frac{1000}{3}R_B - 50\left[\frac{x^3}{3} - 3x^2\right]_6^{10} = 0$$ $$333.33\,R_B - 50\,(33.33 + 36) = 0 \ \Rightarrow\ R_B = \frac{3466.7}{333.33} = 10.40\ \text{kN}\ (\uparrow)$$

Other reactions

$$R_A = 50 - 10.40 = 39.60\ \text{kN}\ (\uparrow)$$ $$M_A = R_B(10) - 50(4) = 10.40(10) - 200 = -96.00\ \text{kNm}$$

The fixed-end moment is 96.00 kNm (hogging).

Bending moment diagram (kNm)

BM is linear in each portion: from 0 at B to $+62.40$ under the load, then falling to $-96.00$ at A. It changes sign at $x = \dfrac{300}{39.6} = 7.58$ m from B (2.42 m from A).

        +62.4
        /\
  B ___/  \___
              \_______ x = 7.58 m (zero)
                       \
                        -96 at A

Answer: $R_B = 10.40$ kN (upward); $R_A = 39.60$ kN; $M_A = 96.00$ kNm (hogging).