--- title: "Truss ILDs for FG and BG" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:truss-ild-fg-bg-2076b" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-fg-bg-2076b site: BCT Engineering --- # Truss ILDs for FG and BG Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2076 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-fg-bg-2076b - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question Draw influence line diagram for members FG and BG of the given truss. [Figure: truss with A (hinge) and E (roller); bottom joints A, F, G, H, E, 4 panels of 15 m = 60 m; top joints B, C, D, with B and D 5 m above and apex C 10 m above the bottom chord; verticals BF, CG, DH and diagonals BG and GD.] ## Short answer FG: always tension, ILD peaks at +2.25 at F and decreases linearly to 0 at A and E. BG: compression only when load is between A and G; ILD is a triangle with 0 at A, −1.581 at F, and 0 at G, zero elsewhere. ## Answer **Data:** bottom chord $A$–$F$–$G$–$H$–$E$, 4 panels of 15 m (span 60 m); $B$, $D$ are 5 m and the apex $C$ is 10 m above the bottom chord; verticals $BF$, $CG$, $DH$; diagonals $BG$ and $GD$. $A$ hinge, $E$ roller. The unit load is taken on the bottom chord (the member forces are zero or linear between panel points). Reactions for a unit load at $x$ from $A$: $R_A = \dfrac{60-x}{60}$, $R_E = \dfrac{x}{60}$. #### Member FG (bottom chord, panel $FG$) Section through $BC$, $BG$ and $FG$ (panel $FG$). Moments about $B$ ($x=15$, $y=5$), where $BC$ and $BG$ meet (lever arm of $FG$ = 5 m): - Load to the right of $G$ (left part): $F_{FG}\times 5 = R_A\times 15 \Rightarrow F_{FG} = 3R_A$ (tension). - Load to the left of $F$ (right part): $F_{FG}\times 5 = R_E\times 45 \Rightarrow F_{FG} = 9R_E$. #### Member BG (diagonal) Use the same section. $AB$ and $BC$ lie on one straight line through $A$ (slope 1/3), so $BC$ and $FG$ meet at $A$; take moments about $A$. The lever arm of $BG$ (line from $B(15,5)$ to $G(30,0)$, length 15.81 m) about $A$ is $$ d = \frac{|15\times(-5) - 5\times 15|}{15.81} = 9.49\ \text{m} $$ - Load to the right of $G$ (left part): the only external force on the left part is $R_A$, acting at $A$, which has zero moment about $A$. So $F_{BG} = 0$. - Load to the left of $F$ (right part): $F_{BG} = -\dfrac{60\,R_E}{9.49} = -6.325\,R_E$ (compression); at $F$, $R_E = 0.25$, so $F_{BG} = -1.581$. - Between $F$ and $G$ the ILD is the straight line from $-1.581$ to 0. #### Ordinates (per unit load; + tension, − compression) | Member | A | F | G | H | E | |---|---|---|---|---|---| | FG | 0.000 | 2.250 | 1.500 | 0.750 | 0.000 | | BG | 0.000 | -1.581 | 0.000 | 0.000 | 0.000 | #### Shapes - **FG:** tension, a triangle with peak $+2.25$ under $F$, falling to $+1.5$ at $G$, $+0.75$ at $H$ and 0 at $E$ (and 0 at $A$). - **BG:** compression, a triangle between $A$ and $G$: 0 at $A$, peak $-1.581$ at $F$, 0 at $G$; zero for load beyond $G$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. To find the force in FG, about which point are moments taken, and why? A. Joint A, where BC and FG meet B. Joint G, where BG and FG meet C. Joint B, where BC and BG meet D. Joint E, the roller Answer: C. Taking moments about B eliminates the unknowns BC and BG, leaving only FG (lever arm 5 m). (Key point: Moment centre for FG) 2. With the load right of G, F_FG equals what in terms of R_A? A. 9 R_A B. 3 R_A (tension) C. 3 R_E D. 0.333 R_A Answer: B. F_FG x 5 = R_A x 15 gives F_FG = 3 R_A, tensile. (Key point: FG expressions) 3. Why is F_BG = 0 for a load to the right of G? A. BG is a zero-force member by joint rule B. The moment of R_A about A is zero, and R_A is the only force on the left part C. The load at G cancels BG D. The lever arm of BG is zero Answer: B. Moments are taken about A, where BC and FG meet; R_A acts at A and has no moment, so BG carries no force. (Key point: Why BG is zero beyond G) 4. Which describes the final ILDs? A. FG compression peak 2.25 at F; BG tension B. FG tension peak +2.25 at F; BG compression peak -1.581 at F, zero from G to E C. FG tension peak at G; BG tension peak at F D. Both are zero everywhere except at F Answer: B. FG ordinates 0, 2.25, 1.5, 0.75, 0; BG is a triangle between A and G with peak -1.581 at F and zero beyond G. 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