--- title: "ILD: bending moment at D, two-span beam" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:ild-bm-d-ab-bc-2" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-bm-d-ab-bc-2 site: BCT Engineering --- # ILD: bending moment at D, two-span beam Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2078 Bhadra, 6 marks - Video: [Influence Line Diagrams for Continuous Beams - Problem No 5 ( Bending Moment at D )](https://www.youtube.com/watch?v=RZvwygTh4E0) — Stan Academy - 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-d-ab-bc-2 - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Draw influence line diagram for bending moment at point D. Find values at interval 2.5 m in span AB and at 5 m interval on span BC. [Figure: two-span beam ABC, A hinged, B and C rollers, AB = 10 m with D at 5 m from A, BC = 20 m.] ## Short answer ILD obtained by adding simple-span M_D and 0.5 times M_B. Ordinates: +2.19 at D, -1.25 at midspan BC (x=20 m); negative peak in BC about -1.28. Zero at supports. ## 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 Two-span beam: AB = 10 m, BC = 20 m (A hinged, B and C rollers), D at 5 m from A. The ILD for $M_D$ is the simple-span ILD for $M_D$ plus the contribution of the continuity moment at B: $$ \text{ILD}(M_D) = \text{ILD}_{\text{simple}}(M_D) + \frac{5}{10}\,\text{ILD}(M_B) $$ with $$ M_B = -\frac{x(L_1^2-x^2)}{2L_1(L_1+L_2)}\ \ (\text{load in AB}),\qquad M_B = -\frac{y(L_2^2-y^2)}{2L_2(L_1+L_2)}\ \ (\text{load in BC},\ y \text{ from C}). $$ The simple-span ordinate is $x/2$ for $x \le 5$ and $(10-x)/2$ for $x \ge 5$. Sample: $x = 5$: simple ordinate $= 2.5$; $M_B = -\dfrac{5(100-25)}{2\times10\times30} = -0.625$; so $M_D = 2.5 + 0.5(-0.625) = 2.1875$. Sample: $x = 15$ ($y = 15$): $M_B = -\dfrac{15(400-225)}{2\times20\times30} = -2.1875$; $M_D = 0.5\times(-2.1875) = -1.094$. #### Ordinates of $M_D$ (2.5 m in AB, 5 m in BC) | x from A (m) | 0 | 2.5 | 5 | 7.5 | 10 | 15 | 20 | 25 | 30 | |---|---|---|---|---|---|---|---|---|---| | $M_D$ | 0 | 1.0547 | 2.1875 | 0.9766 | 0 | -1.0938 | -1.25 | -0.7812 | 0 | #### ILD shape ```text *** * * ** ** ** ** **-------------***----------------------------*** *** ****** ******************* ^ ^ ^ ``` Maximum positive ordinate $+2.19$ at D; the ordinate is zero at B (x = 10 m) and the negative loop in BC reaches about $-1.28$ near x = 18.5 m (the tabulated value at the middle of BC, x = 20 m, is $-1.25$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the ILD for M_D in the two-span beam built? A. Simple-span ILD for M_D plus (5/10) times the ILD of M_B B. Simple-span ILD for M_D minus M_B ILD C. Simple-span ILD only, since D is in span AB D. ILD of M_B multiplied by the span BC length Answer: A. Continuity adds the effect of the support moment M_B, carried to D with the ratio a/L1 = 5/10. (Key point: Composition of ILD) 2. What is the ordinate of M_D when the unit load is at D (x = 5 m)? A. 2.5 B. 2.1875 C. 0.625 D. -1.094 Answer: B. Simple ordinate 2.5 plus 0.5 x (-0.625) = 2.1875. (Key point: Ordinate at D) 3. What is the sign of the M_D ordinates for a load on span BC, and why? A. Positive, because the load is on a span B. Negative, because D is on the left of B C. Negative, because M_B is hogging and the simple-span part is zero D. Zero, because D is not in BC Answer: C. For a load in BC the simple-span part is zero and only 0.5 M_B (hogging) acts. (Key point: Sign in BC) 4. Where is the M_D ordinate zero? A. At A, B and C B. At D only C. At B only D. At the middle of each span Answer: A. Supports are fixed in position, so a load there causes no moment at D; the table gives 0 at x = 0, 10, 30. (Key point: Zero ordinates) ## Related in Theory of Structures II - [ILD: bending moment at B, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-b-two-span-8) — Theory of Structures II, chapter 4 - [ILD: bending moment at fixed support, 10 m beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-fixed-support-10m-1) — Theory of Structures II, chapter 4 - [ILD: moment at section 5, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-section-5-two-span) — Theory of Structures II, chapter 4 - [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: shear at section x-x, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-section-xx-two-span) — Theory of Structures II, chapter 4 - [ILD by focal point method: moment at support 2](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-support-2-focal-point) — Theory of Structures II, chapter 4 ## Related in other subjects - [Suspension bridge triangle ABC; span AB](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Asuspension-bridge-triangle-abc-span-ab) — Surveying I, chapter 7 - [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 - [ILD for bending moment at section D of arch](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-ild-bm-d-given-arch) — Theory of Structures I, chapter 6 - [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