Castigliano: reaction at B, 150 kN overhang

Beam ABC: A fixed, B a roller 4 m from A, overhang BC = 2 m with 150 kN downward at C. EI is constant. The beam has one redundant; take the roller reaction $R_B$ (upward).

 ||==============o=======+ 150 kN
 A (fixed)       B       C
 |<----- 4 ----->|<- 2 ->|

Bending moments (x measured from C towards A, sagging positive)

$$\begin{aligned} CB\ (0 \le x \le 2):\quad & M = -150x, & \frac{\partial M}{\partial R_B} &= 0 \\ BA\ (2 \le x \le 6):\quad & M = -150x + R_B(x-2), & \frac{\partial M}{\partial R_B} &= x-2 \end{aligned}$$

Castigliano's theorem

The deflection at B is zero, so $\dfrac{\partial U}{\partial R_B} = 0$. Let $s = x-2$ (from B):

$$\int_0^4 \left[R_B\,s - 150(s+2)\right]s\,ds = 0$$ $$\frac{R_B (4)^3}{3} - 150\left(\frac{4^3}{3} + 2\cdot\frac{4^2}{2}\right) = 0$$ $$21.333\,R_B - 150(21.333 + 16) = 0 \ \Rightarrow\ R_B = \frac{5600}{21.333} = 262.5\ \text{kN}\ (\uparrow)$$

Other reactions (check)

$$V_A = 150 - 262.5 = -112.5\ \text{kN (i.e. 112.5 kN downward)}$$ $$M_A = R_B(4) - 150(6) = 1050 - 900 = 150\ \text{kNm}$$

Bending moment

In the overhang $M = -150x$, so $M_B = -300$ kNm (hogging). In AB, with $s$ from B, $M = R_B s - 150(s+2) = -300 + 112.5\,s$, which is zero at $s = 2.667$ m from B and equals $+150$ kNm (sagging) at A ($s = 4$).

Answer: $R_B = 262.5$ kN (upward); also $V_A = 112.5$ kN (downward) and $M_A = 150$ kNm.