--- title: "Truss ILDs for BC, CH, HG, bottom chord" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:truss-ild-bc-ch-hg-bottom" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-bc-ch-hg-bottom site: BCT Engineering --- # Truss ILDs for BC, CH, HG, bottom chord Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2080 Chaitra, 10 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-bc-ch-hg-bottom - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question Find ILD for member forces of members BC, CH and HG for the following truss if load moves along the bottom chords. [Figure: truss with A (hinge) at the left, E (roller) at the right; bottom joints A, B, C, D, E with 4 panels of 4 m (16 m); top joints H, G, F at a height of 3 m; verticals HB, GC, FD and diagonals as drawn (AH, HC, CF, FE).] ## Short answer BC: tension, triangular with peak +1.00 at B, 0.667 at C. CH: compression -0.417 at B, tension +0.833 at C, neutral at 5.33 m. HG: compression throughout, peak -1.333 at C. ## Answer **Geometry:** bottom chord $A$–$B$–$C$–$D$–$E$, 4 panels of 4 m (span 16 m); top joints $H$, $G$, $F$ at 3 m height above $B$, $C$, $D$. $A$ hinge, $E$ roller. Diagonals $AH$, $HC$, $CF$, $FE$; verticals $HB$, $GC$, $FD$. The unit load moves along the bottom chord, so ordinates are needed at the panel points only and the ILD is a straight line between them (floor beam action). Reactions: for a unit load at $x$ from $A$, $R_A = \dfrac{16-x}{16}$, $R_E = \dfrac{x}{16}$. #### Method of sections (section through panel $BC$, cutting $HG$, $HC$, $BC$) Diagonal $HC$ has $\sin\theta = 3/5$, $\cos\theta = 4/5$. **Member BC** (take moments about $H$, where $HG$ and $HC$ meet; lever arm 3 m): - Unit load right of the panel ($x\ge 8$, left part considered): $F_{BC}\times 3 = R_A\times 4 \Rightarrow F_{BC} = \tfrac43 R_A$ - Unit load left of the panel ($x\le 4$, right part considered): $F_{BC}\times 3 = R_E\times 12 \Rightarrow F_{BC} = 4R_E = x/4$ **Member HG** (moments about $C$): $F_{HG} = -\dfrac{8R_A}{3}$ for load right of $C$; $F_{HG} = -\dfrac{8R_E}{3}$ for load left of $B$ (top chord in compression). **Member CH** (vertical equilibrium of the left part): $F_{CH}\sin\theta = R_A$ for load right of $C$, so $F_{CH} = \tfrac53 R_A$ (tension); for load left of $B$, $F_{CH} = -\tfrac53 R_E$ (compression). #### Ordinates (per unit load; + tension, − compression) | Member | A | B | C | D | E | |---|---|---|---|---|---| | BC | 0.000 | 1.000 | 0.667 | 0.333 | 0.000 | | CH | 0.000 | -0.417 | 0.833 | 0.417 | 0.000 | | HG | 0.000 | -0.667 | -1.333 | -0.667 | 0.000 | Between $B$ and $C$ the ILDs are straight lines joining the ordinates at $B$ and $C$. #### Shapes - **BC:** tension only; a triangle with peak $+1.00$ at $B$, $0.667$ at $C$, $0$ at $A$ and $E$. - **CH:** compression ($-0.417$ at $B$) for load left of the panel, tension ($+0.833$ at $C$) for load right; a neutral point lies between $B$ and $C$ at $x = 4 + 4\times\frac{0.417}{0.417+0.833} = 5.33$ m from $A$. - **HG:** compression throughout with peak $-1.333$ under $C$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the ILD straight between panel points here? A. The load moves on the bottom chord and is transferred through floor beams at panel points B. The members are all pinned C. The truss is symmetrical D. The reactions are constant Answer: A. Floor beam action makes the ILD linear between panel points. (Key point: Load path) 2. What is the ILD ordinate of BC at B? A. -1.000 B. +1.000 (tension) C. 0.667 D. 0.333 Answer: B. Ordinates are 0, 1.000, 0.667, 0.333, 0 for A to E; BC is tension only. (Key point: Member BC) 3. What is the nature of the ILD of HG? A. Tension throughout B. Tension at B and compression at D C. Compression throughout, with the peak -1.333 under C D. Zero everywhere Answer: C. F_HG = -8R/3 gives a maximum of -1.333 under C. (Key point: Member HG) 4. What does the ILD of CH show? A. Tension only B. Compression only C. No neutral point D. A sign change: compression for load left of the panel BC and tension for load right, neutral at 5.33 m from A Answer: D. Ordinates are -0.417 at B and +0.833 at C, so the line crosses zero at 5.33 m. 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