---
title: "Truss ILDs for six members, load on bottom chord"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:truss-ild-six-members-bottom"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-six-members-bottom
site: BCT Engineering
---

# Truss ILDs for six members, load on bottom chord

Theory of Structures I · Chapter 5: Influence Lines for Simple Structures

- Asked in: 2073 Magh, 12 marks
- Video: [Influence Line For Trusses Top / Upper Chord Member Easiest way to Draw  | ILD & Rolling Load](https://www.youtube.com/watch?v=fuJ2rSQrsc4) — Ganesh Shegar
- 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-six-members-bottom
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw ILD for forces in member $U_1L_1$, $U_2U_3$, $U_2L_3$, $U_2L_2$, $L_2L_3$ and $U_3L_3$ for a given truss, when the load is moving on the bottom chord. [Figure: truss with bottom joints $L_0$ to $L_6$, top joints $U_1$ to $U_5$, 6 panels of length a (6a), height h; $L_0$ hinge, $L_6$ roller.]

## Short answer

ILD shapes: U1L1 triangle with peak +1 at L1; U2U3 compressive, peak -3a/(2h) at L3; U2L3 diagonal, negative left of L2-L3, positive right, peak +d/(2h) at L3; U2L2 positive peak +1/3 at L2, negative peak -1/2 at L3; L2L3 tension, peak +4a/(3h) at L2; U3L3 zero throughout. All zero at supports.

## Answer

**Assumed truss (Pratt type):** bottom joints $L_0$–$L_6$, top joints $U_1$–$U_5$ above $L_1$–$L_5$; 6 panels of length $a$ (span $6a$), height $h$. Verticals $U_iL_i$; diagonals $U_1L_2$, $U_2L_3$, $U_4L_3$, $U_5L_4$; end posts $L_0U_1$ and $L_6U_5$. $L_0$ hinge, $L_6$ roller. Let $d = \sqrt{a^2+h^2}$ be the diagonal length, so $\sin\theta = h/d$. The load travels on the bottom chord, so each ILD is straight between panel points and the ordinate at a panel point equals the force for a unit load placed there.

Reactions for a unit load at $L_k$ ($x = ka$): $R_{L_0} = \dfrac{6-k}{6}$, $R_{L_6} = \dfrac{k}{6}$.

#### Formulas (sections)
- **$U_1L_1$:** $L_1$ has only three members ($L_0L_1$, $L_1L_2$, $U_1L_1$), so it is a hanger: $F = +1$ with the load at $L_1$, 0 otherwise.
- **$U_2U_3$:** cut $U_2U_3$, $U_2L_3$, $L_2L_3$; moments about $L_3$: $F = -\dfrac{M_{L_3}}{h}$ (compression). With the load right of $L_3$: $F = -\dfrac{3a\,R_{L_0}}{h}$; left of $L_3$: $F = -\dfrac{3a\,R_{L_6}}{h}$.
- **$U_2L_3$ (diagonal):** same section, $F = \pm\dfrac{V}{\sin\theta} = \pm\dfrac{d}{h}V$. Load right of $L_3$: $F = +\dfrac{d}{h}R_{L_0}$; left of $L_2$: $F = -\dfrac{d}{h}R_{L_6}$.
- **$L_2L_3$:** same section, moments about $U_2$ ($x = 2a$): $F = \dfrac{M_{U_2}}{h}$ (tension): right of $L_3$: $F = \dfrac{2a\,R_{L_0}}{h}$; left of $L_2$: $F = \dfrac{4a\,R_{L_6}}{h}$.
- **$U_2L_2$ (vertical):** at the unloaded joint $U_2$: $F = -V_{panel\ L_2L_3}$, giving $+R_{L_6}$ for a load at or left of $L_2$ and $-R_{L_0}$ for a load at or right of $L_3$.
- **$U_3L_3$:** no diagonal meets at $U_3$, so this unloaded joint has only three members: the two collinear top-chord members $U_2U_3$, $U_3U_4$ and the vertical $U_3L_3$. Hence $F = 0$ for bottom-chord loading.

#### ILD ordinates (per unit load; + tension, − compression; $d=\sqrt{a^2+h^2}$)
| Member | L1 | L2 | L3 | L4 | L5 |
|---|---|---|---|---|---|
| U1L1 | $1$ | $0$ | $0$ | $0$ | $0$ |
| U2U3 | $- \frac{a}{2 h}$ | $- \frac{a}{h}$ | $- \frac{3 a}{2 h}$ | $- \frac{a}{h}$ | $- \frac{a}{2 h}$ |
| U2L3 | $- \frac{d}{6 h}$ | $- \frac{d}{3 h}$ | $\frac{d}{2 h}$ | $\frac{d}{3 h}$ | $\frac{d}{6 h}$ |
| U2L2 | $\frac{1}{6}$ | $\frac{1}{3}$ | $- \frac{1}{2}$ | $- \frac{1}{3}$ | $- \frac{1}{6}$ |
| L2L3 | $\frac{2 a}{3 h}$ | $\frac{4 a}{3 h}$ | $\frac{a}{h}$ | $\frac{2 a}{3 h}$ | $\frac{a}{3 h}$ |
| U3L3 | $0$ | $0$ | $0$ | $0$ | $0$ |

All ILDs are 0 at $L_0$ and $L_6$.

#### Shapes
- **$U_1L_1$:** triangle with apex $+1$ at $L_1$.
- **$U_2U_3$:** compression, peak $-\dfrac{3a}{2h}$ at $L_3$.
- **$U_2L_3$:** negative to the left of the neutral point between $L_2$ and $L_3$ (peak $-\dfrac{d}{3h}$ at $L_2$), positive to the right (peak $+\dfrac{d}{2h}$ at $L_3$).
- **$U_2L_2$:** positive (peak $+\tfrac13$ at $L_2$), crossing zero between $L_2$ and $L_3$, negative to the right (peak $-\tfrac12$ at $L_3$).
- **$L_2L_3$:** tension, peak $+\dfrac{4a}{3h}$ at $L_2$.
- **$U_3L_3$:** zero throughout.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the ILD of U1L1 for bottom-chord loading?
   A. Triangle with apex +1 at L1, zero elsewhere
   B. Triangle with apex -1 at L1
   C. Constant 1
   D. Peak at L3
   Answer: A. L1 has three members, so U1L1 is a hanger carrying 1 only when the load is at L1. (Key point: Hanger U1L1)

2. How is the force in U2U3 found and what is its nature?
   A. Moments about L3, F = -M_L3/h, compression with peak -3a/(2h) at L3
   B. Moments about U2, tension
   C. Joint method, zero force
   D. Shear, tension with peak at L2
   Answer: A. Taking moments about L3 eliminates the other two cut members; the top chord is in compression. (Key point: Top chord U2U3)

3. Why is U3L3 zero for bottom-chord loading?
   A. It is too short
   B. Joint U3 has two collinear chord members and the vertical, and nothing else, so it is unloaded
   C. It is in compression
   D. It carries the full load
   Answer: B. At U3 two collinear chords and one vertical meet with no external load, so the vertical force is zero. (Key point: Zero-force U3L3)

4. Describe the ILD of diagonal U2L3.
   A. Positive throughout
   B. Negative left of a neutral point between L2 and L3 (peak -d/(3h) at L2), positive to the right (peak +d/(2h) at L3)
   C. Negative throughout
   D. Only nonzero at L2
   Answer: B. Load left of L2 gives compression, load right of L3 gives tension, so the line crosses zero between L2 and L3. (Key point: Diagonal U2L3 sign change)

## Related in Theory of Structures I

- [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 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×
- [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
- [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 and maximum forces under UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-u1u2-u2l2-l2l3-udl) — Theory of Structures I, chapter 5

## Related in other subjects

- [Truss member forces BC, BD, BE, DE, DG, EG](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-bc-bd-be-de-dg-eg-48kn) — Applied Mechanics, chapter 7
- [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 member forces with 25, 50, 10 kN loads](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-25kn-50kn-10kn-hinge-roller) — Applied Mechanics, chapter 7
- [Member forces of truss with 20 and 10 kN loads](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-20kn-10kn-12m-8m) — Applied Mechanics, chapter 7