--- title: "ILD: bending moment at B, two-span beam" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:ild-bm-b-two-span-8" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-bm-b-two-span-8 site: BCT Engineering --- # ILD: bending moment at B, two-span beam Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2078 Kartik, 6 marks - Video: [Influence Line Diagrams for Continuous Beams - Problem No 3 ( Support Moment at B )](https://www.youtube.com/watch?v=K_dti0ceNzk) — 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-b-two-span-8 - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Draw the ILD for bending moment at B. [Figure: two-span beam ABC, A hinged, B and C rollers, AB = 8 m, BC = 10 m.] ## Short answer Moment at B is hogging throughout. Using three-moment equation, ILD ordinates are negative, zero at A, B, C. Peak negative values: -0.667 in AB (at x=4 m), -1.067 in BC (at x=12 m from A). ## 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 ABC (A hinged, B and C rollers), $L_1 = 8$ m, $L_2 = 10$ m, constant $EI$. Take $M_B$ as the redundant; by Muller-Breslau the ILD for $M_B$ is the deflected shape of the beam with a hinge inserted at B and a unit rotation (kink) applied there. Ordinates follow from the three-moment equation with zero moments at A and C. For a unit load at $x$ from A in span AB: $$ M_B = -\frac{x\,(L_1^2 - x^2)}{2L_1(L_1+L_2)} $$ For a unit load at distance $y$ from C in span BC: $$ M_B = -\frac{y\,(L_2^2 - y^2)}{2L_2(L_1+L_2)} $$ Sample: $x = 4$ m: $M_B = -\dfrac{4(64-16)}{2\times8\times18} = -0.667$. $y = 4$ m (x = 14): $M_B = -\dfrac{4(100-16)}{2\times10\times18} = -0.933$. #### Ordinates of $M_B$ | x from A (m) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | |---|---|---|---|---|---|---|---|---|---|---| | $M_B$ | 0 | -0.4167 | -0.6667 | -0.5833 | 0 | -0.8 | -1.0667 | -0.9333 | -0.5333 | 0 | #### ILD shape ```text **-------------------**------------------------** **** ** * ** ***** ***** ** *** *** *** **** ********* ^ ^ ^ ``` The ILD is negative (hogging) throughout and zero at A, B and C. The largest ordinate in AB is about $-0.68$ near $x = 4.6$ m, and in BC about $-1.07$ near $x = 12.2$ m (5.8 m from C); the tabulated value at 12 m is $-1.067$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. By Mueller-Breslau, the ILD for M_B is the deflected shape when... A. Support B is removed B. A hinge is inserted at B and a unit rotation (kink) is applied there C. The beam is cut at B and displaced D. A is fixed Answer: B. A hinge removes moment restraint at B; the unit relative rotation gives the ILD. (Key point: Principle applied) 2. For a unit load at x in AB (L1 = 8, L2 = 10), M_B at x = 4 m is... A. -0.933 B. -0.400 C. -1.067 D. -0.667 Answer: D. M_B = -x(L1^2 - x^2)/(2 L1 (L1+L2)) = -4x48/(2x8x18) = -0.667. (Key point: Ordinate formula) 3. What is the sign of the ILD for M_B? A. Positive in both spans B. Positive in AB, negative in BC C. Negative (hogging) throughout and zero at A, B, C D. Zero in BC Answer: C. A load on either span produces hogging at B; the ordinates vanish at the supports. (Key point: Sign of ILD) 4. Where is the largest ordinate of the M_B ILD and why? A. In AB, because it is shorter B. At B C. At midspan of AB D. In BC (about -1.07), the longer span Answer: D. The ordinate grows with span length; BC (10 m) gives about -1.07 against about -0.68 in AB. 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