--- title: "ILD: moment at middle support, two-span beam" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:ild-moment-mid-support-two-span" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-moment-mid-support-two-span site: BCT Engineering --- # ILD: moment at middle support, two-span beam Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2066 Jestha, 12 marks - Video: [EX 07 | INFLUENCE LINE DIAGRAM | TWO SPAN CONTINUOUS BEAM | INDETERMINATE STRUCTURE | ILD](https://www.youtube.com/watch?v=GZweFbmDFfw) — Structural Funda by:- KheNiSiR - 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-mid-support-two-span - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question A two spanned continuous beam is pinned at the ends. The relative cross-sectional stiffnesses and spans for the left and right beams are 2 EI and 5 EI, 2 m and 6 m respectively. Draw influence line diagram for the moment at the mid support showing ordinates at every 2 m interval. ## Short answer Using three-moment equation with L1/I1=1, L2/I2=1.2 (sum 4.4), M_B = -a(2-a)(2+a)/17.6 for left span, -y(36-y²)/132 for right span. Ordinates at 2m intervals from A: 0m:0, 2m:0, 4m:-0.606, 6m:-0.485, 8m:0. The ILD is hogging everywhere. ## 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. #### Data Beam ABC with pinned ends and a middle support B: $L_1 = 2$ m (stiffness $2EI$), $L_2 = 6$ m (stiffness $5EI$). Ends A and C carry no moment. The ILD for the moment at B is obtained from the three-moment equation with $L_1/I_1 = 2/2 = 1$ and $L_2/I_2 = 6/5 = 1.2$. $$ M_AL_1/I_1 + 2M_B\left(\frac{L_1}{I_1} + \frac{L_2}{I_2}\right) + M_CL_2/I_2 = -(\text{load terms}),\qquad 2\left(1 + 1.2\right) = 4.4 $$ #### Formulas - **Unit load in AB at $a$ from A** ($b = L_1 - a$): load term $= \dfrac{ab(L_1+a)}{I_1L_1}$, so $$ M_B = -\frac{ab(L_1+a)}{I_1L_1\times4.4} = -\frac{a(2-a)(2+a)}{2\times2\times4.4}\ \ (\text{with } I_1 = 2) $$ - **Unit load in BC at $y$ from C**: load term $= \dfrac{y(L_2^2-y^2)}{I_2L_2}$, so $$ M_B = -\frac{y(36-y^2)}{5\times6\times4.4} = -\frac{y(36-y^2)}{132} $$ Sample: $a = 1$: $M_B = -\dfrac{1\times1\times3}{17.6} = -0.1705$. Load 4 m from C ($y = 4$, i.e. $x = 4$ m from A): $M_B = -\dfrac{4\times20}{132} = -0.6061$. (Note: ordinates are given at x measured from A; positions in BC are x = 2 + distance from B.) #### Ordinates of $M_B$ (every 1 m, so the 2 m positions are included) | x from A (m) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |---|---|---|---|---|---|---|---|---|---| | $M_B$ | 0 | -0.1705 | 0 | -0.4167 | -0.6061 | -0.6136 | -0.4848 | -0.2652 | 0 | #### ILD shape ```text ***--------**----------------------------------** ******** ** **** *** **** **** ***** ************ ^ ^ ^ ``` The ILD is negative (hogging) everywhere and zero at A, B and C. The largest ordinate is about $-0.63$ near $x = 4.5$ m (2.5 m from B, in the longer span); the short span AB has a maximum of only about $-0.175$ near $x = 1.15$ m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With pinned ends, what coefficient multiplies M_B in the three-moment equation here? A. 2.2 B. 4.4 C. 1.0 D. 6.0 Answer: B. 2(L1/I1 + L2/I2) = 2(1 + 1.2) = 4.4, since M_A = M_C = 0. (Key point: Three-moment coefficient) 2. What is the nature of the ILD of M_B? A. Positive in AB and negative in BC B. Negative (hogging) everywhere and zero at A, B and C C. Positive everywhere with maximum at B D. Zero at A and C but maximum magnitude at B Answer: B. A unit load anywhere on the continuous beam causes hogging at B, and the ordinate is zero when the load is on a support. (Key point: Sign and zero points) 3. For a unit load at distance y from C in span BC, which formula applies? A. M_B = -y(36-y^2)/132 B. M_B = -y(6-y)/4.4 C. M_B = -y^2/132 D. M_B = +y(36-y^2)/132 Answer: A. The load term y(L2^2 - y^2)/(I2 L2) divided by 4.4 gives -y(36-y^2)/(5 x 6 x 4.4) = -y(36-y^2)/132. (Key point: Load in BC) 4. Where is the largest ordinate of the ILD, and why? A. In AB, because it is the stiffer span B. At support B itself C. In BC (about -0.63 near x = 4.5 m), the longer span D. Equal in both spans Answer: C. The longer span BC gives a much larger ordinate (about -0.63) than the short span AB (about -0.175). (Key point: Where is the maximum) ## Related in Theory of Structures II - [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: 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: 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: 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: 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 ## 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 - [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 - [ILD for BM at F and stress in support BD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-moment-f-stress-bd) — Theory of Structures I, chapter 5 - [Define influence line diagram and its uses](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Adefine-ild-uses) — Theory of Structures I, chapter 5, asked 5×