--- title: "Castigliano: reaction at C, beam with overhang" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:castigliano-support-c-6knm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-support-c-6knm site: BCT Engineering --- # Castigliano: reaction at C, beam with overhang Theory of Structures II · Chapter 2: Force method - Asked in: 2075 Chaitra, 10 marks - Video: [Castigliano's Second Theorem - Deflection at a Point - Continuous Beam with Cantilever](https://www.youtube.com/watch?v=pAhp20WsNNc) — AF Math & Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-support-c-6knm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine the support reaction at support 'C' using Castigliano's theorem. EI = constant throughout. [Figure: beam ABCD, A fixed, a moment applied at B, 2 m from A; support C at 2 m from B; 6 kN/m UDL on CD of 1 m overhang.] ## Short answer Taking roller reaction RC as redundant, set the partial derivative of strain energy with respect to RC to zero and integrate over the beam. RC = 9.9375 kN upward. ## Answer 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. ```text 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) | Range | $M$ | $\partial M/\partial R_C$ | |---|---|---| | DC, $0 \le s \le 1$ | $-3s^2$ | 0 | | CB, $1 \le s \le 3$ | $-6(s-0.5) + R_C(s-1)$ | $s-1$ | | BA, $3 \le s \le 5$ | $-6(s-0.5) + R_C(s-1) - 10$ | $s-1$ | (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) | Point | BM | |---|---| | A | +2.750 | | B (left of couple) | -5.125 | | B (right of couple) | +4.875 | | C | -3.000 | | D | 0 | **Answer: the reaction at the roller C is $R_C = 9.9375$ kN upward ($159/16$ kN).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which condition gives the equation for R_C by Castigliano's theorem? A. dU/dR_C = 0 because C does not settle B. U = 0 C. Sum of moments about C is zero D. dU/dM = 0 at A Answer: A. The deflection at C equals dU/dR_C, which is zero since the support does not settle. (Key point: Condition used) 2. Why is dM/dR_C = 0 in the overhang DC? A. The UDL is zero there B. EI is infinite there C. R_C acts at C and has no moment arm to sections beyond C D. The moment is constant Answer: C. Sections between D and C lie on the free side of C, so M there does not contain R_C. (Key point: Derivative of M) 3. For sections between B and A, how does the clockwise couple at B enter the sagging moment (s from D)? A. It adds +10 kNm B. It reduces it by 10 kNm C. It is ignored D. It multiplies R_C Answer: B. The couple lies on the D-side of the section and is clockwise, so it reduces the sagging moment by 10 kNm. (Key point: Effect of couple) 4. What is R_C? A. 6 kN B. 3.94 kN C. 15 kN D. 9.9375 kN upward Answer: D. (64/3)R_C - 152 - 60 = 0 gives R_C = 212 x 3/64 = 9.9375 kN. 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