--- title: "Castigliano: beam, redundant moment at A" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:castigliano-beam-redundant-moment-a" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-beam-redundant-moment-a site: BCT Engineering --- # Castigliano: beam, redundant moment at A Theory of Structures II · Chapter 2: Force method - Asked in: 2078 Bhadra, 6 marks - Video: [Analysis of Indeterminate beam by Theorem of Least Work or Castiglianis thm](https://www.youtube.com/watch?v=vU11cS32Mxg) — Firoz Nadaf - 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-beam-redundant-moment-a - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Analyze the given loaded beam using Castigliano's theorem. The redundant should be considered as the support moment at A. EI constant. [Figure: beam ABCD, A fixed; 10 kNm moment applied at B (2 m from A); support at C, 2 m from B; 6 kN/m UDL on overhang CD of 1 m.] ## Short answer Release rotation at fixed end A, take clockwise MA as redundant. RA = –(MA+13)/4, ∂M/∂MA = 1 – x/4. Set∫₀⁴ M(1–x/4)dx=0 yields 4MA/3–26/3+5=0, so MA=2.75 kNm. Then RA=–3.94 kN downward, RC=9.94 kN upward. ## Answer Reading of the figure: A fixed; B is 2 m from A where a couple of 10 kNm (taken clockwise) is applied; C is a roller 2 m from B; CD is a 1 m overhang with 6 kN/m. EI is constant. The fixity at A gives one redundant. Release the rotation at A and take the fixed-end moment $M_A$ (clockwise on the beam) as the redundant, so the beam becomes simply supported at A and C with an overhang. The condition is that the slope at A is zero: $$ \frac{\partial U}{\partial M_A} = \theta_A = 0 $$ ```text 10 kNm (cw) 6 kN/m A ||--------B--------C vvvvvvvv D (fixed) ^ (roller) |<-- 2 -->|<-- 2 -->|<-- 1 -->| ``` #### Reaction at A in terms of $M_A$ Moments about C (the overhang load is 6 kN at 0.5 m from C): $$ 4R_A + M_A + 10 + 6(0.5) = 0 \ \Rightarrow\ R_A = -\frac{M_A + 13}{4} $$ $\dfrac{\partial R_A}{\partial M_A} = -\dfrac{1}{4}$ ($R_A$ is positive upward). #### Bending moments (x from A, sagging positive) $$ \begin{aligned} 0 \le x \le 2:\quad M &= M_A + R_A x, & \frac{\partial M}{\partial M_A} &= 1 - \frac{x}{4} \\ 2 \le x \le 4:\quad M &= M_A + R_A x + 10, & \frac{\partial M}{\partial M_A} &= 1 - \frac{x}{4} \end{aligned} $$ The overhang CD does not depend on $M_A$ (its moment comes from the UDL alone), so it adds nothing to the integral. #### Castigliano's theorem $$ \int_0^4 M\left(1-\frac{x}{4}\right)dx = 0 $$ Evaluating each term with $M = M_A - \dfrac{(M_A+13)x}{4} + 10\,[x>2]$: $$ \begin{aligned} M_A\int_0^4\left(1-\tfrac{x}{4}\right)^2dx &= \tfrac{4}{3}M_A \\ -\tfrac{13}{4}\int_0^4 x\left(1-\tfrac{x}{4}\right)dx &= -\tfrac{13}{4}\cdot\tfrac{8}{3} = -\tfrac{26}{3} \\ 10\int_2^4 \left(1-\tfrac{x}{4}\right)dx &= 10(0.5) = 5 \end{aligned} $$ $$ \tfrac{4}{3}M_A - \tfrac{26}{3} + 5 = 0 \ \Rightarrow\ M_A = \tfrac{11}{4} = 2.75\ \text{kNm} $$ #### Other reactions $$ R_A = -\frac{2.75+13}{4} = -3.9375\ \text{kN}\ (\text{downward}), \qquad R_C = 6 - (-3.9375) = 9.9375\ \text{kN}\ (\uparrow) $$ #### Bending moments (kNm) | Point | BM | |---|---| | A | +2.75 (sagging) | | B (left of couple) | -5.125 | | B (right of couple) | +4.875 | | C | -3.000 | | D | 0 | **Answer: $M_A = 2.75$ kNm (clockwise on the beam, provided by the wall), $R_A = 3.938$ kN downward, $R_C = 9.938$ kN upward.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the support moment M_A as the redundant, what condition gives its value? A. The deflection at A is zero: dU/dR_A = 0 B. The moment at C is zero C. The slope at A is zero: dU/dM_A = 0 D. The slope at D is zero Answer: C. A is a fixed end, so rotation at A vanishes; dU/dM_A is the rotation at A. (Key point: Condition used) 2. Why does the overhang CD add nothing to the Castigliano integral? A. Its moment comes from the UDL alone and does not depend on M_A, so dM/dM_A = 0 there B. Because CD is shorter than the other spans C. Because the UDL has no effect on bending D. Because the moment at D is zero, so the whole span is ignored Answer: A. Only parts of the beam whose moment varies with M_A contribute to the integral of M (dM/dM_A). (Key point: Overhang contribution) 3. What does (4/3)M_A - 26/3 + 5 = 0 give? A. M_A = 3.25 kNm B. M_A = 2.75 kNm C. M_A = 6.5 kNm D. M_A = 8.25 kNm Answer: B. (4/3)M_A = 26/3 - 5 = 11/3, so M_A = 11/4 = 2.75 kNm. Then R_A = -3.9375 kN and R_C = 9.9375 kN. 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