--- title: "ILD by focal point method: moment at support 2" subject: "Theory of Structures II" chapter: 4 key: "semester_5/theory_of_structures_ii:ild-moment-support-2-focal-point" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/ild-moment-support-2-focal-point site: BCT Engineering --- # ILD by focal point method: moment at support 2 Theory of Structures II · Chapter 4: Influence line (IL) for continuous beams - Asked in: 2071 Chaitra, 6 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-moment-support-2-focal-point - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-4 ## Question Draw influence line diagram for moment at support 2 of the continuous beam shown in figure below by using the focal point method. Find ordinates at 4 m interval in span 1-2 and at 2 m interval on span 2-3. [Figure: beam 1-2-3, 1 fixed, 2 roller, 3 hinged; span 1-2 = 20 m, span 2-3 = 10 m.] ## Short answer Distribute fixed-end moments using stiffness shares 0.4 to span 2-1 and 0.6 to span 2-3. M_2 is always hogging: maximum -1.728 at x = 12 m in span 1-2, about -0.768 at 24 m in span 2-3, and zero at supports 1, 2 and 3. ## 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 and focal ratios Beam 1-2-3: end 1 fixed, 2 roller, 3 hinged; $L_1 = 20$ m, $L_2 = 10$ m, constant $EI$. For span n with the far end at moment ratio $k$, the unloaded span acts as a rotational spring at the joint with stiffness $S = \dfrac{6EI}{L(2-k)}$, where $k$ is its focal ratio ($|M_{far}/M_{near}|$). - Span 1-2 seen from joint 2 (end 1 fixed): left focal ratio $k = \tfrac12$ (focal point at $L_1/3$ from 1), so $S_{21} = \dfrac{6EI}{20(1.5)} = 0.2EI$. - Span 2-3 seen from joint 2 (end 3 hinged): right focal ratio $k = 0$ (focal point at 3), so $S_{23} = \dfrac{6EI}{10(2)} = 0.3EI$. Distribution of an unbalanced joint moment at 2: share to 2-1 $= 0.2/0.5 = 0.4$, share to 2-3 $= 0.3/0.5 = 0.6$. #### Ordinates of $M_2$ **Unit load in span 1-2 at $x$ from 1** ($a = x$, $b = 20 - x$): the fixed-end moment at 2 is $F_{21} = \dfrac{a^2b}{L_1^2}$, and the part held by span 1-2 is 0.6 of it: $$ M_2 = -0.6\,\frac{x^2(20-x)}{400} $$ **Unit load in span 2-3 at $y$ from 2**: with the hinge at 3 released, the end moment at 2 is $F'_{23} = \dfrac{y(L_2-y)(2L_2-y)}{2L_2^2}$, and the part held is 0.4 of it: $$ M_2 = -0.4\,\frac{y(10-y)(20-y)}{200} $$ Sample: $x = 12$: $M_2 = -0.6\times\dfrac{144\times8}{400} = -1.728$. $y = 2$ (position 22 m): $M_2 = -0.4\times\dfrac{2\times8\times18}{200} = -0.576$. | Distance from 1 (m) | 0 | 4 | 8 | 12 | 16 | 20 | 22 | 24 | 26 | 28 | 30 | |---|---|---|---|---|---|---|---|---|---|---|---| | $M_2$ | 0 | -0.384 | -1.152 | -1.728 | -1.536 | 0 | -0.576 | -0.768 | -0.672 | -0.384 | 0 | (The values above were checked against the exact three-moment solution.) #### ILD shape ```text *****---------------------------*--------------** **** ** **** ***** **** * ***** **** *** ********* ^ ^ ^ ``` The ILD is entirely hogging. The maximum is about $-1.78$ at x = 13.3 m in span 1-2, and about $-0.77$ at 24.2 m in span 2-3; it is zero at 1, 2 and 3. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Span 1-2 (end 1 fixed) has focal ratio and span 2-3 (hinged end 3) has A. 0 and 1/2 B. 1/2 and 1/2 C. 1/2 and 0 D. 0 and 0 Answer: C. Fixed end gives k = 1/2; hinged end gives k = 0. (Key point: Focal ratios) 2. Stiffnesses 0.2EI (span 1-2) and 0.3EI (span 2-3) give shares at 2 of A. 0.6 to 2-1 and 0.4 to 2-3 B. 0.5 each C. 0.2 and 0.3 D. 0.4 to 2-1 and 0.6 to 2-3 Answer: D. Shares are 0.2/0.5 and 0.3/0.5. (Key point: Distribution) 3. For load in span 1-2 the moment at 2 is A. -0.6 x^2(20-x)/400 B. -0.4 x^2(20-x)/400 C. -x^2(20-x)/400 D. -0.6 x(20-x)/20 Answer: A. Span 1-2 holds 0.6 of the fixed-end moment a^2b/L^2, the rest going to span 2-3. (Key point: Load in 1-2) 4. The ILD of M_2 is A. Sagging in span 2-3 B. Hogging throughout, zero at 1, 2 and 3 C. Positive in span 1-2 D. Zero in span 2-3 Answer: B. All tabulated ordinates are negative or zero. 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