--- title: "ILD: bending moment at fixed support, 10 m beam" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:ild-bm-fixed-support-10m-1" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-bm-fixed-support-10m-1 site: BCT Engineering --- # ILD: bending moment at fixed support, 10 m beam Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2068 Baishakh, 10 marks - 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/ild-bm-fixed-support-10m-1 - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Draw influence line diagram for bending moment at the fixed support of the beam and obtain ordinates at each 1.25 m interval. [Figure: propped cantilever, fixed at the left, roller at the right, span 10 m, EI constant.] ## Short answer For a 10 m propped cantilever, MA = x(L-x)(2L-x)/(2L^2), hogging negative. Ordinates per unit load at 1.25 m intervals are 0, -1.0254, -1.6406, -1.9043, -1.875, -1.6113, -1.1719, -0.6152, 0; maximum about -1.925 at x = 4.22 m. ## Answer Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam. #### Method Propped cantilever AB, span $L = 10$ m, A fixed, B roller, constant $EI$. The bending moment at the fixed support A is the reaction moment (hogging). Muller-Breslau: replace the fixed support by a hinge and apply a unit rotation at A; the deflected shape of the beam is the ILD of $M_A$. #### Derivation of the ordinates Take the roller reaction $R_B$ as the redundant. For a unit load at $x$ from A: 1. Deflection at B of the released cantilever (fixed at A) under the unit load: $\delta_{Bx} = \dfrac{x^2(3L-x)}{6EI}$. 2. Deflection at B under a unit upward force at B: $f_{BB} = \dfrac{L^3}{3EI}$. 3. Compatibility $\delta_{Bx} = R_Bf_{BB}$ gives $R_B = \dfrac{x^2(3L-x)}{2L^3}$. 4. Moment at A (hogging): $M_A = x\cdot 1 - R_BL = x - \dfrac{x^2(3L-x)}{2L^2} = \dfrac{x(L-x)(2L-x)}{2L^2}$. Sample: $x = 5$: $R_B = \dfrac{25\times25}{2000} = 0.3125$, $M_A = 5 - 0.3125\times10 = 1.875$ (hogging). $x = 3.75$: $R_B = \dfrac{14.0625\times26.25}{2000} = 0.1846$, $M_A = 3.75 - 1.846 = 1.904$. #### Ordinates (1.25 m interval, hogging = negative) | x from A (m) | 0 | 1.25 | 2.5 | 3.75 | 5 | 6.25 | 7.5 | 8.75 | 10 | |---|---|---|---|---|---|---|---|---|---| | $M_A$ | 0 | -1.0254 | -1.6406 | -1.9043 | -1.875 | -1.6113 | -1.1719 | -0.6152 | 0 | #### ILD shape ```text **--------------------------------------------*** ** ***** **** ***** ***** ******* **************** ^ ^ ``` The maximum ordinate is about $-1.925$ at $x = 4.22$ m ($0.42L$). A concentrated load $P$ at $x = 5$ m gives $M_A = 1.875P$ hogging; a UDL $w$ over the full span gives $M_A = w\times$ area $= wL^2/8 = 12.5w$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. By the Muller-Breslau principle, how is the ILD of the moment at the fixed support A obtained? A. Replace the fixed support by a hinge and give a unit rotation at A; the deflected shape is the ILD. B. Remove the roller and apply a unit upward displacement at B. C. Cut the beam at A and give a unit vertical shift between the faces. D. Replace the fixed support by a roller and apply a unit load at A. Answer: A. Releasing the moment restraint (hinge) and imposing a unit rotation gives the deflected shape that is the ILD of M_A. (Key point: Muller-Breslau) 2. In the derivation, which condition gives the roller reaction R_B? A. Sum of vertical forces equals zero. B. Deflection at B of the released cantilever under the unit load equals R_B times the flexibility f_BB. C. Moment at B equals zero. D. Slope at A equals the slope at B. Answer: B. R_B is the redundant; compatibility of deflection at B (delta_Bx = R_B f_BB with f_BB = L^3/3EI) gives R_B = x^2(3L-x)/2L^3. (Key point: Redundant and compatibility) 3. Which expression gives the hogging moment at A for a unit load at x? A. M_A = R_B L + x B. M_A = x - R_B L = x(L-x)(2L-x)/2L^2 C. M_A = x(L-x)/L D. M_A = R_B x - L Answer: B. Taking moments about A for the cantilever: unit load gives x, the upward R_B gives R_B L, so M_A = x - R_B L. (Key point: Moment expression) 4. For L = 10 m, what is M_A for a unit load at mid-span, and the effect of a full-span UDL w? A. 1.875 hogging; M_A = wL^2/8 = 12.5w B. 1.875 hogging; M_A = wL^2/2 = 50w C. 2.5 hogging; M_A = 12.5w D. 0.3125 hogging; M_A = 12.5w Answer: A. At x = 5, R_B = 0.3125 and M_A = 5 - 3.125 = 1.875 (hogging); UDL times ILD area gives wL^2/8 = 12.5w. (Key point: Key ordinates and use) ## Related in Theory of Structures II - [ILD: moment at support B, fixed-propped beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-b-propped-2l-l) — Theory of Structures II, chapter 4 - [ILD: fixed-end moment of propped cantilever](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-reaction-moment-fixed-propped-10m) — Theory of Structures II, chapter 4, asked 2× - [ILD: moment at middle support, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-mid-support-two-span) — Theory of Structures II, chapter 4 - [ILD: fixed-end reaction, propped cantilever 5 m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-fixed-end-reaction-propped-5m) — Theory of Structures II, chapter 4, asked 2× - [ILD: reaction at B, propped cantilever 10 m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-reaction-b-propped-10m-2m) — Theory of Structures II, chapter 4, asked 2× - [ILD: bending moment at D, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-d-ab-bc-2) — Theory of Structures II, chapter 4 ## Related in other subjects - [Overhang beam: ILD, max positive/negative moment at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aoverhang-ild-moment-d-three-loads) — Theory of Structures I, chapter 5 - [Propped cantilever with UDL, two elements](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Apropped-cantilever-two-elements) — Computational Techniques in Civil Engineering, chapter 4 - [ILD at mid-span: BM and SF under given loads](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-mid-span-bm-sf-50-10-30) — Theory of Structures I, chapter 5 - [ILDs for BM and SF at 10 m, 25 m beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-beam-bm-sf-10m-25m) — Theory of Structures I, chapter 5, asked 2×