---
title: "Truss: max forces in CF and BC under live load"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:truss-ild-cf-bc-live-load-28"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-cf-bc-live-load-28
site: BCT Engineering
---

# Truss: max forces in CF and BC under live load

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

- Asked in: 2072 Asoj, 12 marks
- Video: [Chapter 6-Influence Lines for Trusses](https://www.youtube.com/watch?v=GzqjDVsMFyQ) — Maen
- 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-cf-bc-live-load-28
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Determine the maximum force in the member CF and BC of the truss as shown due to a live load of 28 kN/m longer than the span passing over the truss. [Figure: truss with A (hinge) and E (roller); bottom joints A, B, C, D, E, 4 panels of 6 m = 24 m; top joints F, G, H, with F and H at 3 m and apex G at 4 m above the bottom chord; verticals FB and HD, diagonals FC, CH as drawn.]

## Short answer

BC: ILD tension only, maximum force 504.0 kN when entire span loaded. CF: ILD has positive and negative areas of ±4.025 m; max tension 112.7 kN (load from neutral point 9.60 m to E), max compression 112.7 kN (load from A to 9.60 m).

## Answer

**Data:** bottom chord $A$–$B$–$C$–$D$–$E$, 4 panels of 6 m (span 24 m); top joints $F$, $H$ 3 m and apex $G$ 4 m above the bottom chord; members $AF$, $FG$, $GH$, $HE$ (top chord), verticals $FB$, $GC$, $HD$ and diagonals $FC$, $CH$. $A$ hinge, $E$ roller. The live load of 28 kN/m is longer than the span, so it can be placed on any part of the span (the loaded length is chosen to suit the ILD). Reactions for a unit load at $x$: $R_A = \dfrac{24-x}{24}$, $R_E = \dfrac{x}{24}$.

#### Member BC (bottom chord, panel $BC$)
Section through $FG$, $FC$ and $BC$; moments about $F$ (lever arm 3 m):

- Load right of $C$ (left part): $F_{BC} = \dfrac{6\,R_A}{3} = 2R_A$.
- Load left of $B$ (right part): $F_{BC} = \dfrac{18\,R_E}{3} = 6R_E$.

#### Member CF (diagonal)
Same section; $FG$ (slope $\tfrac16$) meets the bottom chord at $O$ ($x = -12$ m). The lever arm of $CF$ about $O$ is $10.73$ m.

- Load right of $C$: $F_{CF} = \dfrac{12\,R_A}{10.73} = +1.118\,R_A$ (tension).
- Load left of $B$: $F_{CF} = -\dfrac{36\,R_E}{10.73} = -3.354\,R_E$ (compression).

#### Ordinates (per unit load; + tension, − compression)
| Member | A | B | C | D | E |
|---|---|---|---|---|---|
| CF | 0.000 | -0.839 | 0.559 | 0.280 | 0.000 |
| BC | 0.000 | 1.500 | 1.000 | 0.500 | 0.000 |

$CF$ changes sign between $B$ and $C$ at $x = 9.60$ m from $A$.

#### Maximum forces for w = 28 kN/m (load of any length)
Force $= w\times$ (area of ILD loaded).

| Member | Positive area (m) | Negative area (m) | w x area (kN) |
|---|---|---|---|
| BC | 18.000 | 0.000 | 504.0 (T) |
| CF | 4.025 | -4.025 | 112.7 (T) / -112.7 (C) |

**Answer: member $BC$: maximum tension = 504.0 kN (whole span loaded; $BC$ is never in compression). Member $CF$: maximum tension = 112.7 kN (load from $x = 9.60$ m to $E$); maximum compression = 112.7 kN (load from $A$ to $x = 9.60$ m).**

## Check yourself

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

1. For a 28 kN/m load longer than the span, how is maximum force found?
   A. Load whole span always
   B. Load only the ILD portions of the required sign: w x area
   C. Place at the peak ordinate only
   D. Use w x span
   Answer: B. The loaded length is chosen to cover only the portion of the ILD with the required sign. (Key point: Meaning of load longer than span)

2. For BC, moments about F with lever arm 3 m. With the load right of C, F_BC equals:
   A. 6 R_A / 3 = 2 R_A
   B. 18 R_E / 3
   C. R_A / 3
   D. 12 R_A / 3
   Answer: A. The left part's only force is R_A at 6 m from F horizontally, giving F x 3 = R_A x 6, so F = 2 R_A. (Key point: BC moment centre)

3. What is the maximum force in BC?
   A. 112.7 kN
   B. 504.0 kN tension with the whole span loaded
   C. 252 kN compression
   D. 28 kN
   Answer: B. All ordinates are positive, area 18 m, so 28 x 18 = 504 kN tension; BC is never in compression. (Key point: BC maximum)

4. Why can CF be both in tension and compression?
   A. Its ILD changes sign at x = 9.60 m, with equal areas 4.025 m
   B. It is a zero-force member
   C. The load reverses
   D. Its ILD is all positive
   Answer: A. The ILD is -0.839 at B and +0.559 at C, so it crosses zero; 28 x 4.025 = 112.7 kN each way. (Key point: CF reversal)

## Related in Theory of Structures I

- [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 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
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- [Truss: max forces under long UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-max-forces-members-1-2-3-4) — 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 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

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