Castigliano: moment at C of propped cantilever

Assumptions: A is a hinge at $x=0$, B a roller at $x=2$ m, the 30 kNm couple acts clockwise at D ($x=4$ m) and C is the fixed end at $x=6$ m. EI is constant. The beam has four vertical/rotational reactions ($R_A,R_B,R_C,M_C$) and two equilibrium equations, so it is indeterminate to the 2nd degree. Take $R_A$ and $R_B$ (upward) as redundants. Castigliano's theorem for redundants that do not move: $\partial U/\partial R_A=0$ and $\partial U/\partial R_B=0$.

Bending moments (x from A, sagging +)

Compatibility equations

$$\frac{\partial U}{\partial R_A}=\frac{1}{EI}\int M\frac{\partial M}{\partial R_A}dx=0:\quad \int_0^2R_Ax\cdot x\,dx+\int_2^4[R_Ax+R_B(x-2)]\,x\,dx+\int_4^6[R_Ax+R_B(x-2)+30]\,x\,dx=0$$

which simplifies to

$$72R_A+37.333R_B+300=0$$

Similarly

$$\frac{\partial U}{\partial R_B}=0:\quad \int_2^4[R_Ax+R_B(x-2)](x-2)dx+\int_4^6[R_Ax+R_B(x-2)+30](x-2)dx=0$$ $$37.333R_A+21.333R_B+180=0$$

Solution

$$R_A=2.250\ \text{kN}\uparrow,\qquad R_B=-12.375\ \text{kN}\ (\text{negative: acts downward})$$ $$R_C=-(R_A+R_B)=10.125\ \text{kN}\uparrow$$

Moment at the fixed end C

$$M_C=R_A(6)+R_B(4)+30=13.500+(-49.500)+30=-6.00\ \text{kNm}$$

The negative sign shows hogging (tension at the top) at C.

Check: $\sum V = R_A+R_B+R_C = 2.250-12.375+10.125 = 0$ ✓. The same $M_C$ is obtained by an independent stiffness analysis.

Answer: $|M_C|=6.00$ kNm, hogging (for a clockwise 30 kNm couple); $R_A=2.25$ kN, $R_B=12.375$ kN (downward), $R_C=10.125$ kN.