Castigliano: fixed moment, propped cantilever UDL

Propped cantilever AB: A fixed, B a roller, span $L = 6$ m, UDL $w = 5$ kN/m. $E = 232$ kN/mm$^2$ and $I = 112.5\times10^6$ mm$^4$ give $EI = 232\times10^6\times1.125\times10^{-4} = 26100$ kNm$^2$, but the reactions of a single-span propped cantilever do not depend on $EI$.

Take the roller reaction $R_B$ as the redundant. With $x$ measured from B:

$$M = R_B\,x - \frac{wx^2}{2},\qquad \frac{\partial M}{\partial R_B} = x$$

The deflection at B is zero:

$$\frac{\partial U}{\partial R_B} = \frac{1}{EI}\int_0^6\left(R_Bx - 2.5x^2\right)x\,dx = 0$$ $$\frac{R_B(6)^3}{3} - \frac{2.5(6)^4}{4} = 0 \ \Rightarrow\ 72R_B = 810 \ \Rightarrow\ R_B = 11.25\ \text{kN}$$

(This is $3wL/8 = 3(5)(6)/8 = 11.25$ kN.)

Moment at the fixed support A

$$M_A = R_B(6) - \frac{5(6)^2}{2} = 67.5 - 90 = -22.5\ \text{kNm}$$

The fixed-end moment is 22.5 kNm (hogging, $= wL^2/8 = 5(36)/8$).

The vertical reaction at A is $30 - 11.25 = 18.75$ kN. The maximum sagging moment is $\dfrac{9wL^2}{128} = 12.66$ kNm at $x = \dfrac{3L}{8} = 2.25$ m from B.

Answer: $M_A = 22.5$ kNm (hogging); $R_B = 11.25$ kN, $R_A = 18.75$ kN.