--- title: "Castigliano: propped cantilever, moment and rotation" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:castigliano-propped-30knm-50kn-moment-rotation" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-propped-30knm-50kn-moment-rotation site: BCT Engineering --- # Castigliano: propped cantilever, moment and rotation Theory of Structures II · Chapter 2: Force method - Asked in: 2076 Asoj, 10 marks - Video: [Castigliano's theorem numerical | Find rotation at B | PU 2021, TU 2075,2076,2071 | most important](https://www.youtube.com/watch?v=uRARqSlQ-gQ) — Sudip Dhakal - 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-30knm-50kn-moment-rotation - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine the moment at the fixed support and the rotation at the roller support of a propped cantilever beam of span 10 m loaded with a uniformly distributed load of 30 kN/m on its whole span and a point load of 50 kN at the centre using Castigliano's theorem. ## Short answer Taking the roller reaction as redundant and applying Castigliano’s theorem, the fixed-end moment is 468.75 kNm (hogging) and the roller rotation is 781.25/EI rad (anticlockwise). ## Answer The beam AB (span 10 m) is fixed at A and has a roller at B. Loads: 30 kN/m over the whole span and 50 kN at the centre. EI is constant. Take the roller reaction $R_B$ as the redundant. ```text 50 kN v 30 kN/m vvvvvvvvvvvvvvvvvvvvvvvvvv ||==========+===========o A (fixed) 5 m B (roller, R_B) |<------------ 10 m ---->| ``` #### Bending moment (x measured from B) $$ M = R_B\,x - 15x^2 - 50\langle x-5\rangle, \qquad \frac{\partial M}{\partial R_B} = x $$ #### Redundant from Castigliano The deflection at B is zero: $$ \frac{\partial U}{\partial R_B} = \frac{1}{EI}\int_0^{10} M\,x\,dx = 0 $$ $$ \int_0^{10}\left(R_B x^2 - 15x^3\right)dx - 50\int_5^{10}(x-5)x\,dx = 0 $$ $$ \frac{1000}{3}R_B - 37500 - 5208.33 = 0 \ \Rightarrow\ R_B = 128.125\ \text{kN} $$ (Check: UDL alone gives $3wL/8 = 112.5$ kN and the central load gives $5P/16 = 15.625$ kN; their sum is 128.125 kN.) #### Moment at the fixed support $$ V_A = 350 - 128.125 = 221.875\ \text{kN} $$ $$ M_A = R_B(10) - 30\cdot\frac{10^2}{2} - 50(5) = 128.125(10) - 1500 - 250 = -468.75\ \text{kNm} $$ The fixed-end moment is **468.75 kNm (hogging)**. #### Rotation at the roller B Apply a dummy couple $M'$ at B. Then $\partial M/\partial M' = 1$ at the sections (the extra term caused by the change of $R_B$ integrates to zero because $\int M\,x\,dx = 0$), so $$ \theta_B = \frac{\partial U}{\partial M'}\bigg|_{M'=0} = \frac{1}{EI}\int_0^{10} M\,dx $$ $$ \int_0^{10} M\,dx = R_B\frac{10^2}{2} - 15\frac{10^3}{3} - 50\frac{5^2}{2} = 128.125(50) - 5000 - 625 = 781.25 $$ $$ \theta_B = \frac{781.25}{EI}\ \text{rad (anticlockwise)} $$ **Answer: fixed-end moment $M_A = 468.75$ kNm (hogging); rotation at the roller $\theta_B = \dfrac{781.25}{EI}$ rad.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the redundant and the condition that gives it? A. R_B at the roller; the deflection at B is zero, so dU/dR_B = 0 B. M_A at the fixed end; slope at A equals zero C. R_B; the moment at B is zero D. V_A; the shear at A is zero Answer: A. Removing the roller leaves a cantilever; the roller prevents deflection at B. (Key point: Redundant and condition) 2. What is R_B and how can it be checked? A. 112.5 kN; the point load adds nothing B. 128.125 kN; UDL alone gives 3wL/8 = 112.5 kN and the point load gives 5P/16 = 15.625 kN C. 350 kN, the total load D. 75 kN, half of the UDL Answer: B. By superposition the propped-cantilever reactions add up to 128.125 kN. (Key point: Value of R_B) 3. What is the moment at the fixed support? A. 375 kNm hogging B. 1750 kNm hogging C. 468.75 kNm hogging D. 250 kNm hogging Answer: C. M_A = 128.125 x 10 - 30 x 10^2/2 - 50 x 5 = -468.75 kNm. The value 375 would be for the UDL alone. (Key point: Fixed-end moment) 4. Why is a dummy couple M' applied at B, and what is theta_B? A. To make the beam determinate; theta_B = 0 B. To replace the roller; theta_B = 468.75/EI C. To measure the deflection at B; theta_B = 128.125/EI D. There is no real couple at B, so M' is added to differentiate and then set to zero; theta_B = 781.25/EI Answer: D. dU/dM' at M' = 0 gives the rotation: (1/EI) times the integral of M dx = 781.25/EI. 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