--- title: "ILD: moment at support B, fixed-propped beam" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:ild-moment-b-propped-2l-l" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-moment-b-propped-2l-l site: BCT Engineering --- # ILD: moment at support B, fixed-propped beam Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2076 Asoj, 6 marks - Video: [Muller Breslau Principle Proof - Influence Line Diagrams for Indeterminate Structures](https://www.youtube.com/watch?v=4C6kwqFJEso) — Ekeeda - 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-moment-b-propped-2l-l - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Draw Influence Line diagram for the moment at support B of a propped cantilever beam as shown. Plot ordinates at 0.50 times span length. [Figure: two-span beam ABC; A fixed, B interior support, C roller; AB = 2L, BC = L, EI constant.] ## Short answer Using moment distribution, M_B is hogging throughout. At mid-AB (x=L) ordinate is -0.15L; at mid-BC (x=2.5L) ordinate is -0.075L. 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 Beam ABC: A fixed, B and C supports, $AB = 2L$, $BC = L$, constant $EI$. Use the moment-distribution (stiffness) form: rotational stiffness at B of AB is $4EI/2L = 2EI/L$ (far end fixed) and of BC is $3EI/L$ (far end C is a roller, moment zero). So the distribution factors at B are $DF_{BA} = 2/5 = 0.4$ and $DF_{BC} = 3/5 = 0.6$. **Unit load in AB at $x$ from A:** $M_{BA}^F = \dfrac{x^2(2L-x)}{(2L)^2}$; support moment $M_B = (1 - 0.4)M_{BA}^F$: $$ M_B = -0.6\,\frac{x^2(2L-x)}{4L^2} $$ **Unit load in BC at $y$ from B:** the fixed-end moment at B with C released is $\dfrac{y(L-y)(2L-y)}{2L^2}$, and $M_B = (1-0.6)$ of it: $$ M_B = -0.4\,\frac{y(L-y)(2L-y)}{2L^2} $$ Sample: $x = L$: $M_B = -0.6\times\dfrac{L^2\cdot L}{4L^2} = -0.15L$. $y = 0.5L$: $M_B = -0.4\times\dfrac{0.5\cdot0.5\cdot1.5}{2}L = -0.075L$. #### Ordinates (at 0.25 and 0.5 times each span length) | Position | Load at | Ordinate of $M_B$ | |---|---|---| | A | 0L | $0L$ | | 0.25 AB | 0.5L | $-0.0563L$ | | 0.5 AB (L from A) | 1L | $-0.15L$ | | 0.75 AB | 1.5L | $-0.1687L$ | | B | 2L | $0L$ | | 0.25 BC | 2.25L | $-0.0656L$ | | 0.5 BC | 2.5L | $-0.075L$ | | 0.75 BC | 2.75L | $-0.0469L$ | | C | 3L | $0L$ | #### ILD shape ```text *****---------------------------*--------------** **** ** **** ***** **** * ***** **** *** ********* ^ ^ ^ ``` The ILD is hogging throughout; the largest ordinate in AB is about $-0.178L$ near $x = 1.33L$, and in BC about $-0.077L$ near $0.42L$ from B (mid-span value $-0.075L$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With AB = 2L (far end fixed) and BC = L (far end roller), constant EI, the DFs at B are A. DF_BA = 0.4, DF_BC = 0.6 B. DF_BA = 0.6, DF_BC = 0.4 C. Both 0.5 D. DF_BA = 2/3, DF_BC = 1/3 Answer: A. Stiffness of AB is 4EI/2L = 2EI/L, of BC 3EI/L; shares 2/5 and 3/5. (Key point: Distribution factors) 2. Why is the stiffness of BC at B taken as 3EI/L? A. Far end C is fixed B. Far end C is a roller, so moment there is zero C. BC is shorter than AB D. Because EI is constant Answer: B. A far end that is simply supported gives 3EI/L rather than 4EI/L. (Key point: Stiffness of BC) 3. What is the general nature of the ILD for M_B? A. Positive in AB, negative in BC B. Positive throughout C. Hogging (negative) throughout, zero at A, B and C D. Negative in AB, zero in BC Answer: C. All tabulated ordinates are negative or zero. 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