--- title: "Truss ILDs for L0U1, L0L1, U2L2" subject: "Theory of Structures I" chapter: 5 key: "semester_4/theory_of_structures_i:truss-ild-l0u1-l0l1-u2l2-chords" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-l0u1-l0l1-u2l2-chords site: BCT Engineering --- # Truss ILDs for L0U1, L0L1, U2L2 Theory of Structures I · Chapter 5: Influence Lines for Simple Structures - Asked in: 2078 Chaitra, 9 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-l0u1-l0l1-u2l2-chords - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5 ## Question Draw influence line diagram for members $L_0U_1$, $L_0L_1$ and $U_2L_2$ when unit load moves on the chords. [Figure: truss with bottom joints $L_0$ to $L_6$ and top joints $U_1$ to $U_5$, 6 panels of 5 m = 30 m, height 5 m, $L_0$ hinge, $L_6$ roller.] ## Short answer L0U1 ILD: always compression, ordinate = −1.414 R_L0, max −1.179 at L1. L0L1 always tension, ordinate = +R_L0, max +0.833 at L1. U2L2: force = −shear in panel L2L3; bottom-chord loading gives +0.333 at L2, −0.500 at L3; top-chord gives −0.667 at U2. ## Answer **Assumed truss (figure not available):** Pratt truss, 6 panels of 5 m (span 30 m), height 5 m; verticals $U_iL_i$ and diagonals $U_1L_2$, $U_2L_3$, $U_4L_3$, $U_5L_4$; end posts $L_0U_1$ and $L_6U_5$ at 45°. $L_0$ hinge, $L_6$ roller. Reaction for unit load at $x$ from $L_0$: $R_{L_0} = (30-x)/30$. #### Method - **$L_0U_1$** (end post): joint $L_0$, vertical equilibrium: $F\sin45^\circ = -R_{L_0}$, so $F_{L_0U_1} = -1.414\,R_{L_0}$ (always compression). - **$L_0L_1$:** horizontal equilibrium at $L_0$: $F_{L_0L_1} = -F_{L_0U_1}\cos45^\circ = +R_{L_0}$ (always tension). - **$U_2L_2$** (vertical): the section through $U_2U_3$, $U_2L_3$ and $L_2L_3$ gives the diagonal $F_{U_2L_3} = \sqrt2\,V$, where $V$ is the shear in panel $L_2L_3$ (left forces upward positive). At the unloaded joint $U_2$ the vertical balances the diagonal: $F_{U_2L_2} = -F_{U_2L_3}/\sqrt2 = -V$. For a unit load at $L_2$: $V = \tfrac23 - 1 = -\tfrac13$, so $F_{U_2L_2} = +0.333$. For a unit load on the top chord at $U_2$, the load itself is added at the joint, which gives $-0.667$. #### Ordinates when the unit load moves on the bottom chord (+ tension) | Member | L0 | L1 | L2 | L3 | L4 | L5 | L6 | |---|---|---|---|---|---|---|---| | L0U1 | 0.000 | -1.179 | -0.943 | -0.707 | -0.471 | -0.236 | 0.000 | | L0L1 | 0.000 | 0.833 | 0.667 | 0.500 | 0.333 | 0.167 | 0.000 | | U2L2 | 0.000 | 0.167 | 0.333 | -0.500 | -0.333 | -0.167 | 0.000 | #### Ordinates when the unit load moves on the top chord | Member | U1 | U2 | U3 | U4 | U5 | |---|---|---|---|---|---| | L0U1 | -1.179 | -0.943 | -0.707 | -0.471 | -0.236 | | L0L1 | 0.833 | 0.667 | 0.500 | 0.333 | 0.167 | | U2L2 | 0.167 | -0.667 | -0.500 | -0.333 | -0.167 | #### Shapes (bottom-chord loading) - **$L_0U_1$:** always compression. 0 at $L_0$ (load directly on the support), peak $-1.179$ at $L_1$, then a straight line decreasing to 0 at $L_6$. Beyond $L_1$ the ordinate is $-1.414\,R_{L_0}$. - **$L_0L_1$:** always tension. 0 at $L_0$, peak $+0.833$ at $L_1$, then falling linearly to 0 at $L_6$ (ordinate $=R_{L_0}$). - **$U_2L_2$:** $+0.167$ at $L_1$, $+0.333$ at $L_2$, $-0.5$ at $L_3$, then $-0.333$, $-0.167$ and 0. The sign changes between $L_2$ and $L_3$, at $x = 10 + 5\times\frac{0.333}{0.833} = 12$ m from $L_0$, so the vertical is stressed in both tension and compression (the part of the ILD with a negative area is larger). - With **top-chord loading**, the ordinates of $L_0U_1$ and $L_0L_1$ are the same linear functions of $R_{L_0}$ (taken at the top joints), while $U_2L_2$ takes $-0.667$ at $U_2$, since the load acts directly on the vertical. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the nature of the ILD of L0U1? A. Always compression, F = -1.414 R_L0 B. Tension C. Both tension and compression D. Zero Answer: A. Vertical equilibrium at L0 gives F sin 45 = -R_L0. (Key point: End post) 2. What is the force in L0L1 compared to R_L0? A. F = -R_L0 B. F = +R_L0, always tension C. F = 1.414 R_L0 D. F = 0 Answer: B. Horizontal equilibrium at L0 gives F = -F_L0U1 cos 45 = R_L0. (Key point: Bottom chord) 3. What is the sign change position of U2L2 (bottom chord loading)? A. At L1 B. At L4 C. About 12 m from L0, between L2 and L3 D. No sign change Answer: C. Ordinates are +0.333 at L2 and -0.5 at L3, so the zero crossing is at 10 + 5(0.333/0.833) = 12 m. (Key point: Vertical U2L2) 4. How does U2L2 differ with a load at the top joint U2? A. It is +0.667 B. It is zero C. It is unchanged at 0.333 D. The ordinate is -0.667 because the load acts directly on the vertical Answer: D. The load at joint U2 is added into the vertical. (Key point: Top chord loading) ## Close variants set in other exams - [Truss ILDs for LC, LK, LD, lower chord](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-lc-lk-ld-lower) — 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 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 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, 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 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 - [Truss ILDs for FG and BC; max forces from 100 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-fg-bc-100kn-roll) — 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 - [Frame AFD, SFD, BMD (30 kN/m, 50 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30knm-50knm) — Strength of Materials, chapter 1