Castigliano: fixed-end moment, 15 kN load

Method: the propped cantilever is indeterminate to the 1st degree. Take the roller reaction $R_B$ (upward) as the redundant. B does not move, so by Castigliano's theorem $\partial U/\partial R_B=0$.

Data: $L=5$ m, 15 kN at 2 m from the fixed end A (so 3 m from B), EI constant.

Bending moment (x measured from B towards A, sagging +)

• $0\le x\le3$: $M=R_Bx$, $\;\partial M/\partial R_B=x$
• $3\le x\le5$: $M=R_Bx-15(x-3)$, $\;\partial M/\partial R_B=x$

Condition $\partial U/\partial R_B=0$

$$\int_0^3R_Bx\cdot x\,dx+\int_3^5\left[R_Bx-15(x-3)\right]x\,dx=0$$ $$R_B\left[\frac{x^3}{3}\right]_0^5-15\int_3^5(x^2-3x)dx=0 \;\Rightarrow\; \frac{125}{3}R_B-15(8.667)=0 \;\Rightarrow\; R_B=\frac{130\times3}{125}=3.12\ \text{kN}\uparrow$$

Fixed-end moment

At A ($x=5$):

$$M_A=R_B(5)-15(5-3)=15.60-30=-14.40\ \text{kNm}$$

The negative sign means hogging (tension at the top).

Check with the standard formula: $M_A=\dfrac{Pab(L+b)}{2L^2}=\dfrac{15\times2\times3\times(5+3)}{2\times25}=14.40$ kNm ✓. Also $R_A=15-3.12=11.88$ kN.

Answer: fixed-end moment $M_A=14.40$ kNm (hogging); $R_B=3.12$ kN, $R_A=11.88$ kN.