--- title: "ILD: moment at section 5, two-span beam" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:ild-bm-section-5-two-span" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-bm-section-5-two-span site: BCT Engineering --- # ILD: moment at section 5, two-span beam Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2074 Chaitra, 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-section-5-two-span - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Draw the influence line for bending moment at Section 5 of a two span continuous beam as shown in figure below. Given ordinate at 2 m interval. [Figure: two-span beam, each span 4@2 m = 8 m, sections 1 to 6 marked at 2 m intervals, middle support B.] ## Short answer Using superposition: M5 = simple-span M5 + 0.5 M_B. At section (12 m) ordinate is +1.625; in AB at 4 m it is -0.375; 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. #### Reading of the figure The beam has two equal spans of $4@2 = 8$ m, hinged at A and on rollers at B and C. Section 5 is taken as the middle section of span BC (3 sections per span, numbered 1-3 in AB and 4-6 in BC), that is 12 m from A (4 m from B and 4 m from C). #### Method The ILD for the moment at a section in BC is the ILD of the simple span BC plus the effect of the support moment $M_B$: $$ \text{ILD}(M_5) = \text{ILD}_{\text{simple}} + \frac{4}{8}\,\text{ILD}(M_B) $$ Here $M_B = -\dfrac{x(L_1^2 - x^2)}{2L_1(L_1+L_2)}$ for a load in AB at $x$ from A, and $M_B = -\dfrac{y(L_2^2 - y^2)}{2L_2(L_1+L_2)}$ for a load in BC at $y$ from C. With $L_1 = L_2 = 8$ m: $M_B = -\dfrac{x(64-x^2)}{256}$. Sample: load at the section ($y = 4$): simple ordinate $= \dfrac{4\times4}{8} = 2.0$; $M_B = -\dfrac{4(64-16)}{256} = -0.75$; $M_5 = 2.0 + 0.5(-0.75) = 1.625$. Sample: load at 4 m from A: $M_B = -0.75$, $M_5 = 0.5(-0.75) = -0.375$. #### Ordinates (2 m interval) | x from A (m) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | |---|---|---|---|---|---|---|---|---|---| | $M_5$ | 0 | -0.2344 | -0.375 | -0.3281 | 0 | 0.6719 | 1.625 | 0.7656 | 0 | #### ILD shape ```text *** ** *** *** *** *** *** ******----------------*****--------------------** **************** ^ ^ ^ ``` Positive (sagging) in BC with a peak of $+1.625$ under the section, and a small negative loop in AB (about $-0.385$ near x = 4.6 m; $-0.375$ at x = 4 m), zero at A, B and C. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Section 5 is taken as A. The middle of span BC, 12 m from A B. The middle support B C. The middle of span AB D. The right support C Answer: A. Three sections per span: 1-3 in AB, 4-6 in BC, so 5 is mid-BC. (Key point: Section location) 2. ILD(M5) = A. Simple-span ILD of BC - ILD(M_B) B. Simple-span ILD of BC + (4/8) ILD(M_B) C. Simple-span ILD of AB + ILD(M_B) D. (4/8) ILD(M_B) only Answer: B. The support moment M_B at the end of BC is carried to the section with ratio 4/8. (Key point: Composition) 3. Maximum ordinate under Section 5 is A. +2.0 B. +0.75 C. +1.625 D. -0.375 Answer: C. 2.0 + 0.5 x (-0.75) = 1.625. (Key point: Peak ordinate) 4. For a load in AB, M5 is A. Negative, equal to 0.5 M_B B. Positive C. Zero because the section is in BC D. Equal to M_B Answer: A. Only the continuity moment acts: at x = 4, 0.5 x (-0.75) = -0.375. (Key point: Load in AB) ## Related in Theory of Structures II - [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: moment at section x-x, continuous beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-section-xx-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: 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 - [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 ## Related in other subjects - [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 - [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 - [Cantilever 3 m: bending stress 25 mm below top](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acantilever-3m-udl-20kn-t-section) — Strength of Materials, chapter 7