--- title: "Truss ILDs for AB, BC, BG, lower chord" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:truss-ild-ab-bc-bg-lower-chord" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-ab-bc-bg-lower-chord site: BCT Engineering --- # Truss ILDs for AB, BC, BG, lower chord Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2068 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-ab-bc-bg-lower-chord - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question Draw influence line diagrams for the forces in members AB, BC and BG of the truss. The load moves in the lower chord of the truss. [Figure: truss with A (hinge) and E (roller); bottom joints A, F, G, H, E, 4 panels of 15 m = 60 m; B and D 5 m above the bottom chord, apex C 10 m above; verticals BF, CG, DH and diagonals BG, GD.] ## Short answer ILD ordinates: AB compression, 0 at A, peak −2.372 at F, then to 0 at E; BC compression, peak −1.581 at G; BG compression in panel AF, peak −1.581 at F, zero for loads beyond G. All shapes are piecewise linear. ## Answer **Data:** bottom chord $A$–$F$–$G$–$H$–$E$, 4 panels of 15 m (span 60 m); $B$, $D$ are 5 m and $C$ is 10 m above the bottom chord; verticals $BF$, $CG$, $DH$; diagonals $BG$, $GD$. $A$ hinge, $E$ roller. The load moves on the lower chord. Reactions: $R_A = \dfrac{60-x}{60}$, $R_E = \dfrac{x}{60}$. $AB$ and $BC$ both have slope $\tfrac13$ (so $A$, $B$, $C$ are collinear). #### Member AB Joint $A$: vertical equilibrium, $F_{AB}\sin\alpha = -R_A$ with $\sin\alpha = \dfrac{5}{15.81} = 0.316$. So $F_{AB} = -3.162\,R_A$ (compression); the load at $A$ goes straight to the support, so the ordinate at $A$ is 0. #### Member BC Section through $BC$, $BG$, $FG$; moments about $G$ (lever arm of $BC$: $\dfrac{30}{\sqrt{10}} = 9.487$ m). - Load right of $G$: $F_{BC} = -\dfrac{30R_A}{9.487} = -3.162R_A$. - Load left of $F$: $F_{BC} = -3.162R_E$. #### Member BG Moments about $A$ (where $BC$ and $FG$ meet; lever arm of $BG$ = 9.49 m). - Load right of $G$: $F_{BG} = 0$ (the left part has only $R_A$, which passes through $A$). - Load left of $F$: $F_{BG} = -\dfrac{60R_E}{9.49} = -6.325R_E$. #### Ordinates (per unit load; + tension, − compression) | Member | A | F | G | H | E | |---|---|---|---|---|---| | AB | 0.000 | -2.372 | -1.581 | -0.791 | 0.000 | | BC | 0.000 | -0.791 | -1.581 | -0.791 | 0.000 | | BG | 0.000 | -1.581 | 0.000 | 0.000 | 0.000 | #### Shapes - **AB:** compression, 0 at $A$, peak $-2.372$ at $F$, falling linearly to 0 at $E$. - **BC:** compression, peak $-1.581$ under $G$; $-0.791$ under $F$ and $H$. - **BG:** compression, a triangle between $A$ and $G$ with peak $-1.581$ at $F$; zero for load beyond $G$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is F_AB = -3.162 R_A? A. AB is a tension member B. Joint A vertical equilibrium: F_AB sin(alpha) = -R_A with sin(alpha) = 0.316 C. AB is horizontal D. Moments about G Answer: B. At joint A the only vertical forces are R_A and the vertical component of AB, so F_AB = -R_A/0.316 = -3.162 R_A (compression). (Key point: Member AB from joint A) 2. What is the peak ordinate of the BC influence line, under G (x = 30 m)? A. -3.162 B. -0.791 C. -2.372 D. -1.581 Answer: D. At G, R_A = 0.5, so F_BC = -3.162 x 0.5 = -1.581 (compression). (Key point: BC peak ordinate) 3. Why is F_BG = 0 for a load to the right of G? A. Moments about A: the left part has only R_A which passes through A, so BG carries nothing B. Because BG is a zero-force member by joint rule always C. Because the load goes directly to E D. Because BG is vertical Answer: A. BC and FG meet at A, the moment centre. For a load right of G the left part has only R_A, whose moment about A is zero, so F_BG = 0. (Key point: BG moment centre) 4. What is the AB ordinate under F (x = 15 m)? A. -1.581 B. -2.372 C. -0.791 D. -3.162 Answer: B. R_A at F = 45/60 = 0.75, so F_AB = -3.162 x 0.75 = -2.372. 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