---
title: "Truss ILDs for FG and BC; max forces from 100 kN"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:truss-ild-fg-bc-100kn-roll"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-fg-bc-100kn-roll
site: BCT Engineering
---

# Truss ILDs for FG and BC; max forces from 100 kN

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

- Asked in: 2071 Magh, 5 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-bc-100kn-roll
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw influence line diagram for forces in member FG and BC of the truss shown in figure below and determine maximum forces in these members when a single concentrated load 100 kN rolls over the span of the truss. [Figure: truss with A (hinge) and E (roller); bottom joints A, B, C, D, E with 4 panels of 4 m; apex G at a height of 5 m; top joints F and H; verticals FB, GC, HD and diagonals as drawn.]

## Short answer

ILD ordinates for FG: 0, −0.472, −0.943, −0.472, 0; for BC: 0, 1.2, 0.8, 0.4, 0. The maximum forces for a 100 kN rolling load are FG = −94.3 kN (compression) with load at C, and BC = 120 kN (tension) with load at B.

## Answer

**Data:** bottom chord $A$–$B$–$C$–$D$–$E$, 4 panels of 4 m (span 16 m); apex $G$ 5 m above the bottom chord. Assumed: $F$ and $H$ at 2.5 m (so $AFG$ and $GHE$ are straight rafters); verticals $FB$, $GC$, $HD$; diagonals $FC$ and $CH$. $A$ hinge, $E$ roller. Reactions: $R_A = \dfrac{16-x}{16}$, $R_E = \dfrac{x}{16}$. The unit load moves along the bottom chord.

#### Member BC (bottom chord, panel $BC$)
Section through $FG$, $FC$, $BC$; moments about $F$ (4, 2.5): lever arm 2.5 m.

- Load right of $C$: $F_{BC} = \dfrac{4R_A}{2.5} = 1.6R_A$.
- Load left of $B$: $F_{BC} = \dfrac{12R_E}{2.5} = 4.8R_E$.

#### Member FG (top chord)
Same section; moments about $C$. The lever arm of $FG$ (line $y = 0.625x$) about $C$ is $\dfrac{5}{\sqrt{1+0.625^2}} = 4.24$ m.

- Load right of $C$: $F_{FG} = -\dfrac{8R_A}{4.24} = -1.887R_A$ (compression).
- Load left of $B$: $F_{FG} = -\dfrac{8R_E}{4.24} = -1.887R_E$.

#### Ordinates (per unit load; + tension, − compression)
| Member | A | B | C | D | E |
|---|---|---|---|---|---|
| FG | 0.000 | -0.472 | -0.943 | -0.472 | 0.000 |
| BC | 0.000 | 1.200 | 0.800 | 0.400 | 0.000 |

Between $B$ and $C$ both ILDs are straight lines.

```text
 ILD BC: 0 / 1.2 at B \ 0.8 at C ... 0.4 at D ... 0 at E
 ILD FG: 0 / -0.47 at B / -0.94 at C \ -0.47 at D \ 0
```

#### Maximum forces for a single 100 kN rolling load
A single load gives its maximum effect at the largest ordinate:

- **FG:** largest ordinate $-0.943$ at $C$ (load at the centre): $F_{FG} = 100\times(-0.943) = -94.3$ kN (compression).
- **BC:** largest ordinate $+1.2$ at $B$: $F_{BC} = 100\times 1.2 = +120$ kN (tension).

**Answer: $FG_{max} = 94.3$ kN (C) with the load at $C$; $BC_{max} = 120$ kN (T) with the load at $B$.**

## Check yourself

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

1. Where does a single rolling load give maximum effect?
   A. At the largest ordinate of the ILD
   B. At mid-span always
   C. At the support
   D. At the smallest ordinate
   Answer: A. Effect = P x ordinate, so the maximum is at the largest ordinate. (Key point: Single load maximum)

2. What are the FG ordinate and maximum force?
   A. -0.943 at C; -94.3 kN compression
   B. +1.2 at B; 120 kN
   C. -0.472 at B; 47 kN
   D. +0.943 at C; tension
   Answer: A. 100 x (-0.943) = -94.3 kN (compression) with the load at C. (Key point: FG maximum)

3. Where is the BC ILD largest and what is the max force?
   A. At C, 80 kN
   B. At B, +1.2, giving 120 kN tension
   C. At D, 40 kN
   D. At A, 160 kN
   Answer: B. ILD ordinates 1.2, 0.8, 0.4 so max at B: 100 x 1.2 = 120 kN tension. (Key point: BC maximum)

## Related in Theory of Structures I

- [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
- [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: max forces in CF and BC under live load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-cf-bc-live-load-28) — Theory of Structures I, chapter 5
- [Truss ILDs for FG and BG](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-fg-bg-2076b) — 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 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

## Related in other subjects

- [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
- [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
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