--- title: "Castigliano: reaction at B, 20 kN/m beam" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:castigliano-reaction-b-20knm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-reaction-b-20knm site: BCT Engineering --- # Castigliano: reaction at B, 20 kN/m beam Theory of Structures II · Chapter 2: Force method - Asked in: 2078 Kartik, 6 marks - Video: [Castigliano’s Theorem Made Easy: Solve within Minutes](https://www.youtube.com/watch?v=sur6mZ_66ak) — Earthquake and Structures - 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-reaction-b-20knm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Find the reaction at support B using Castigliano's method. [Figure: beam ABC with 20 kN/m UDL over its full length; AB = 6 m, BC = 10 m; A hinged, B roller.] ## Short answer Choose RB as redundant, express RA = 160 – 0.625 RB. Integrate M·∂M/∂RB over AB and BC using Castigliano’s theorem, set ∂U/∂RB = 0, yielding 75 RB – 15800 = 0. Hence RB = 210.67 kN upward. ## Answer Reading of the figure: a continuous beam ABC with $AB = 6$ m and $BC = 10$ m carrying 20 kN/m over the whole length, A hinged, B and C on rollers (a beam with only A and B supported would be determinate, so C is taken as a roller). EI is constant. Take $R_B$ as the redundant. ```text 20 kN/m vvvvvvvvvvvvvvvvvvvvvvvvvv A o--------o B-----------o C |<-- 6 -->|<---- 10 ---->| ``` #### Reactions in terms of $R_B$ Moments about C: $R_A(16) + R_B(10) = 20(16)(8)$ $$ R_A = 160 - 0.625\,R_B $$ #### Bending moments (x from A) $$ \begin{aligned} 0 \le x \le 6:\quad M &= R_A x - 10x^2, & \frac{\partial M}{\partial R_B} &= -0.625x \\ 6 \le x \le 16:\quad M &= R_A x + R_B(x-6) - 10x^2, & \frac{\partial M}{\partial R_B} &= 0.375x - 6 \end{aligned} $$ #### Castigliano's theorem The support B does not settle, so $\partial U/\partial R_B = 0$: $$ \int_0^6 M\,\frac{\partial M}{\partial R_B}\,dx + \int_6^{16} M\,\frac{\partial M}{\partial R_B}\,dx = 0 $$ Substituting $R_A$ and integrating gives $$ 75.000\,R_B - 15800.00 = 0 $$ $$ R_B = \frac{15800.00}{75.000} = 210.67\ \text{kN} $$ #### Other reactions $$ R_A = 160 - 0.625(210.67) = 28.33\ \text{kN}, \qquad R_C = 320 - 210.67 - 28.33 = 81.00\ \text{kN} $$ Check: $M_B = 28.33(6) - 20(6)(3) = -190$ kNm, which agrees with the three moment equation $2M_B(16) = -\tfrac{20}{4}(6^3 + 10^3)$. **Answer: $R_B = 210.67$ kN (upward).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is dU/dR_B set to zero? A. Because R_B is the largest reaction B. Support B does not settle, so the deflection at B (equal to dU/dR_B) is zero C. Because the strain energy is minimum only when R_B = 0 D. Because the moment at B is zero Answer: B. By Castigliano's theorem dU/dR_B is the deflection at B in the direction of R_B, and a rigid support allows none. (Key point: Castigliano condition) 2. For 0 <= x <= 6, with R_A = 160 - 0.625 R_B, what is dM/dR_B? A. x B. -x C. -0.625x D. 0.375x - 6 Answer: C. In AB, M = R_A x - 10x^2 and only R_A depends on R_B, so dM/dR_B = x (dR_A/dR_B) = -0.625x. (Key point: Moment derivative) 3. What does the solution of 75 R_B - 15800 = 0 give? A. R_B = 210.67 kN upward, with R_A = 28.33 kN and R_C = 81.00 kN B. R_B = 160 kN, the simple-span reaction C. R_B = 28.33 kN upward D. R_B = 15800 kN Answer: A. R_B = 15800/75 = 210.67 kN. R_A then follows from 160 - 0.625 R_B, and R_C from the total load of 320 kN. 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