--- title: "Truss ILDs for LC, LK, LD, lower chord" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:truss-ild-lc-lk-ld-lower" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-lc-lk-ld-lower site: BCT Engineering --- # Truss ILDs for LC, LK, LD, lower chord Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2070 Bhadra, 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-lc-lk-ld-lower - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question Draw influence line diagram for the members LC, LK and LD when the load moves in the lower chord of the given truss as shown in figure. [Figure: truss with bottom joints A, B, C, D, E, F, G (6 panels of 6 m = 36 m), top joints N, M, L, K, J, I, H, height 8 m; A hinge, G roller.] ## Short answer Influence ordinates: LK is entirely compression, peak −1.125 at D; LD changes sign, maximum tension 0.625 at D, maximum compression −0.417 at C; LC tension peak 0.333 at C, compression peak −0.5 at D, sign change at 14.4 m from A. ## Answer **Assumed truss (Pratt type):** bottom joints $A$–$G$, top joints $N, M, L, K, J, I, H$ directly above them, 6 panels of 6 m (span 36 m), height 8 m; verticals at every joint; diagonals $NB$, $MC$, $LD$, $JD$, $IE$, $HF$ (sloping down towards the centre). $A$ hinge, $G$ roller. The load moves on the lower chord. Reactions: $R_A = \dfrac{36-x}{36}$, $R_G = \dfrac{x}{36}$. The diagonal $LD$ has length $10$ m, $\sin\theta = 0.8$. #### Member LK (top chord, panel $CD$) Section through $LK$, $LD$ and $CD$ (panel $CD$); $LD$ and $CD$ meet at $D$ ($x = 18$ m), so take moments about $D$ (lever arm 8 m): $F_{LK} = -\dfrac{M_D}{8}$. - Load right of $D$: $F_{LK} = -\dfrac{18R_A}{8} = -2.25R_A$; load left of $C$: $F_{LK} = -\dfrac{18R_G}{8} = -2.25R_G$ (compression). #### Member LD (diagonal) Same section: the panel shear is carried by the diagonal, $F_{LD} = \pm\dfrac{V}{0.8}$. - Load right of $D$: $F_{LD} = +\dfrac{R_A}{0.8} = 1.25R_A$ (tension). - Load left of $C$: $F_{LD} = -1.25R_G$ (compression). #### Member LC (vertical) At the unloaded top joint $L$ the vertical balances the vertical component of the diagonal: $F_{LC} = -0.8\,F_{LD}$. #### Ordinates (per unit load; + tension, − compression) | Member | A | B | C | D | E | F | G | |---|---|---|---|---|---|---|---| | LC | 0.000 | 0.167 | 0.333 | -0.500 | -0.333 | -0.167 | 0.000 | | LK | 0.000 | -0.375 | -0.750 | -1.125 | -0.750 | -0.375 | 0.000 | | LD | 0.000 | -0.208 | -0.417 | 0.625 | 0.417 | 0.208 | 0.000 | #### Shapes - **LK:** compression only, triangle with peak $-1.125$ under $D$ (centre). - **LD:** compression (peak $-0.417$ at $C$) for load left of the panel, tension (peak $+0.625$ at $D$) for load to the right; the neutral point is at $x = 12+6\times\frac{0.417}{0.417+0.625} = 14.4$ m from $A$. - **LC:** tension for load on the left (peak $+0.333$ at $C$), compression to the right (peak $-0.5$ at $D$); the sign change lies between $C$ and $D$ at $x = 12 + 6\times\frac{0.333}{0.333+0.5} = 14.4$ m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. To get the force in top chord LK (panel CD) by the method of sections, about which point are moments taken? A. Joint L B. Joint C C. Joint D, where LD and CD meet D. Joint K Answer: C. LD and CD meet at D, so moments about D eliminate them and leave only LK with an 8 m lever arm. (Key point: Moment centre for LK) 2. With the moment centre D at x = 18 m, lever arm 8 m, what is the peak ordinate of the LK ILD? A. -0.750 B. -1.125 C. -2.250 D. -0.500 Answer: B. M_D per unit load at D = 18 x 18/36 = 9, so F_LK = -9/8 = -1.125 (compression). (Key point: Peak ordinate of LK) 3. Why is F_LD = V/0.8 for the diagonal LD? A. The panel shear is carried only by the diagonal, whose vertical component is F sin(theta) with sin(theta) = 0.8 B. The diagonal carries the chord moment C. Because the diagonal is 8 m long D. Because the vertical LC is zero Answer: A. The chords in the cut are nearly horizontal, so the vertical shear is resisted by the diagonal alone: F_LD x 0.8 = V. (Key point: Diagonal LD from panel shear) 4. At the unloaded top joint L, how is F_LC related to F_LD? A. F_LC = F_LD B. F_LC = 0.8 F_LD with same sign C. F_LC = -0.8 F_LD (balances the vertical component of the diagonal) D. F_LC = 1.25 F_LD Answer: C. Joint L has no load; the vertical LC must balance the vertical component 0.8 F_LD of the diagonal, so F_LC = -0.8 F_LD. (Key point: Vertical LC from joint L) ## Close variants set in other exams - [Truss ILDs for L0U1, L0L1, U2L2](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-l0u1-l0l1-u2l2-chords) — Theory of Structures I, chapter 5, asked 2× ## Related in Theory of Structures I - [Truss ILDs for six members, load on bottom chord](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-six-members-bottom) — Theory of Structures I, chapter 5 - [Truss ILDs for AB, BC, BG, lower chord](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-ab-bc-bg-lower-chord) — Theory of Structures I, chapter 5 - [Truss ILDs for BC, CH, HG, bottom chord](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-bc-ch-hg-bottom) — Theory of Structures I, chapter 5 - [Truss ILDs for bc, hg, df, upper chord](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-bc-hg-df-upper-chord) — Theory of Structures I, chapter 5 - [Truss ILDs for DK, DJ, DE](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-dk-dj-de) — Theory of Structures I, chapter 5 - [Truss ILDs for BC and BG; max force in BC](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-bc-bg-live-load-8m) — Theory of Structures I, chapter 5 ## Related in other subjects - [Truss: temperature decrease in vertical members](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-temp-decrease-vertical-ab-ef) — Theory of Structures II, chapter 2 - [Truss: bar forces due to load and temperature](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-load-and-temperature-9kn-18kn) — Theory of Structures II, chapter 2 - [Truss: force method, EA and 2EA members](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-force-method-ea-2ea-50kn) — Theory of Structures II, chapter 2 - [Forces in all members of truss with 10.8 kN loads](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-10-8kn-d-support) — Applied Mechanics, chapter 7