--- title: "Castigliano: prop reaction with BMD" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:castigliano-propped-50kn-reaction-b-bmd" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-propped-50kn-reaction-b-bmd site: BCT Engineering --- # Castigliano: prop reaction with BMD Theory of Structures II · Chapter 2: Force method - Asked in: 2071 Shrawan, 10 marks - Video: [EX 01 | ENERGY PRINCIPLE | PROPPED CANTILEVER BEAM | REACTIONS USING MINIMUM ENERGY CONCEPT](https://www.youtube.com/watch?v=D0GEsufQ6a4) — Structural Funda by:- KheNiSiR - 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-propped-50kn-reaction-b-bmd - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine the reaction at B of the propped cantilever beam shown in figure below using Castigliano's theorem. Also draw the bending moment diagram. [Figure: beam AB, A fixed, B roller, 50 kN point load at 4 m from A, span 10 m (4 m + 6 m).] ## Short answer Setting derivative of strain energy with respect to R_B to zero gives R_B = 10.40 kN upward. Then R_A = 39.60 kN and fixed-end moment M_A = 96 kNm hogging; maximum sagging moment under load is 62.4 kNm. ## Answer Propped cantilever AB: A fixed, B a roller, span 10 m, 50 kN at 4 m from A (6 m from B). EI is constant. Take $R_B$ as the redundant and measure $x$ from B. ```text 50 kN v B o-----------+---------|| A (fixed) |<--- 6 m --->|<- 4 m ->| ``` #### Bending moments $$ \begin{aligned} 0 \le x \le 6:\quad & M = R_Bx, & \frac{\partial M}{\partial R_B} &= x \\ 6 \le x \le 10:\quad & M = R_Bx - 50(x-6), & \frac{\partial M}{\partial R_B} &= x \end{aligned} $$ #### Castigliano's theorem ($\delta_B = 0$) $$ \int_0^{10}R_Bx^2\,dx - 50\int_6^{10}(x-6)\,x\,dx = 0 $$ $$ \frac{1000}{3}R_B - 50\left[\frac{x^3}{3} - 3x^2\right]_6^{10} = 0 $$ $$ 333.33\,R_B - 50\,(33.33 + 36) = 0 \ \Rightarrow\ R_B = \frac{3466.7}{333.33} = 10.40\ \text{kN}\ (\uparrow) $$ #### Other reactions $$ R_A = 50 - 10.40 = 39.60\ \text{kN}\ (\uparrow) $$ $$ M_A = R_B(10) - 50(4) = 10.40(10) - 200 = -96.00\ \text{kNm} $$ The fixed-end moment is **96.00 kNm (hogging)**. #### Bending moment diagram (kNm) | Section | BM | |---|---| | B | 0 | | Under the load (6 m from B) | 62.40 (sagging) | | A (fixed end) | -96.00 (hogging) | BM is linear in each portion: from 0 at B to $+62.40$ under the load, then falling to $-96.00$ at A. It changes sign at $x = \dfrac{300}{39.6} = 7.58$ m from B (2.42 m from A). ```text +62.4 /\ B ___/ \___ \_______ x = 7.58 m (zero) \ -96 at A ``` **Answer: $R_B = 10.40$ kN (upward); $R_A = 39.60$ kN; $M_A = 96.00$ kNm (hogging).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which integral expresses delta_B = 0 here? A. Integral of R_B x dx from 0 to 10 = 0 B. Integral of R_B x^2 dx from 0 to 10 minus 50 times integral of (x-6)x dx from 6 to 10 = 0 C. Integral of 50(x-6) dx = 0 D. Integral of M dx = 0 only Answer: B. U derivative: M (dM/dR_B) = [R_B x - 50(x-6)] x integrated over the beam. (Key point: Equation for the redundant) 2. What is R_B? A. 50 kN B. 39.6 kN C. 6 kN D. 10.40 kN Answer: D. 333.33 R_B = 3466.7 gives R_B = 10.40 kN, upward. (Key point: Redundant value) 3. What is M_A? A. 96 kNm hogging B. 200 kNm hogging C. 62.4 kNm sagging D. 104 kNm sagging Answer: A. M_A = 10.40(10) - 50(4) = -96 kNm. (Key point: Fixed-end moment) 4. Where is the maximum sagging moment and its value? A. At A, 96 kNm B. At B, 62.4 kNm C. Under the load, 62.4 kNm D. At mid-span, 50 kNm Answer: C. Under the load (6 m from B), M = 10.40 x 6 = 62.4 kNm, sagging. 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