Castigliano: reaction at C, beam with overhang

Reading of the figure: A fixed; a couple of 10 kNm (taken clockwise) is applied at B, 2 m from A; C is a roller support 2 m from B; the overhang CD (1 m) carries 6 kN/m. EI is constant. The beam is indeterminate to the first degree: take $R_C$ (upward) as the redundant.

          10 kNm (cw)        6 kN/m
 A ||======B===========C vvvvvv D
 fixed   |<- 2 ->|<- 2 ->|<- 1 ->|

Bending moments (s measured from the free end D; sagging positive)

(The clockwise couple at B, lying on the D-side of any section between B and A, reduces the sagging moment by 10 kNm.)

Castigliano's theorem

C does not settle, so $\dfrac{\partial U}{\partial R_C} = 0$:

$$\int_1^3 \left[-6(s-0.5) + R_C(s-1)\right](s-1)\,ds + \int_3^5 \left[-6(s-0.5) + R_C(s-1) - 10\right](s-1)\,ds = 0$$

Term by term over $1 \le s \le 5$:

$$\begin{aligned} R_C\int_1^5 (s-1)^2\,ds &= \tfrac{64}{3}R_C \\ -6\int_1^5 (s-0.5)(s-1)\,ds &= -152 \\ -10\int_3^5 (s-1)\,ds &= -60 \end{aligned}$$ $$\tfrac{64}{3}R_C - 152 - 60 = 0 \ \Rightarrow\ R_C = \frac{212\times3}{64} = 9.9375\ \text{kN}\ (\uparrow)$$

Other reactions

• Vertical: $R_A = 6 - 9.9375 = -3.9375$ kN, that is 3.9375 kN downward at A.
• Fixed-end moment: $M_A = -6(4.5) + 9.9375(4) - 10 = 2.75$ kNm (sagging side, clockwise reaction on the beam).

Bending moment values (kNm)

Answer: the reaction at the roller C is $R_C = 9.9375$ kN upward ($159/16$ kN).