---
title: "Truss ILDs for BC and BG; max force in BC"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:truss-ild-bc-bg-live-load-8m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-bc-bg-live-load-8m
site: BCT Engineering
---

# Truss ILDs for BC and BG; max force in BC

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

- Asked in: 2072 Magh, 12 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-bc-bg-live-load-8m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw influence line diagram for forces in members BC and BG and determine maximum force in member BC when uniformly distributed load 6 kN/m of length 8 m moves. [Figure: truss with A (hinge) and E (roller); bottom joints A, H, G, F, E, 4 panels of 6 m = 24 m; top joints B, C, D with B and D at 4 m and apex C at 6 m above the bottom chord; verticals BH and DF, diagonals BG, CG, GD as drawn.]

## Short answer

ILD for BC: compression, ordinates -0.527 at H, -1.054 at G. BG: -0.751 at H, +0.300 at G, zero crossing. Maximum force in BC: place 8 m UDL symmetrically about center, area under ILD = 7.027 m, force = 6 × 7.027 = 42.16 kN compression.

## Answer

**Data:** bottom chord $A$–$H$–$G$–$F$–$E$, 4 panels of 6 m (span 24 m); $B$, $D$ are 4 m and the apex $C$ is 6 m above the bottom chord; verticals $BH$, $DF$, $CG$; diagonals $BG$, $GD$. $A$ hinge, $E$ roller. The live load moves along the bottom chord; ILDs are straight between panel points.

Reactions: $R_A = \dfrac{24-x}{24}$, $R_E = \dfrac{x}{24}$.

#### Member BC (top chord, panel $HG$)
Section through $BC$, $BG$ and $HG$. The line $BC$ ($x - 3y + 6 = 0$) has slope $\tfrac13$; the lever arm of $BC$ about $G$ (where $BG$ and $HG$ meet) is $\dfrac{|12+6|}{\sqrt{10}} = 5.692$ m.

- Load right of $G$ (left part): $F_{BC} = -\dfrac{12\,R_A}{5.692} = -2.108\,R_A$ (compression).
- Load left of $H$ (right part): $F_{BC} = -\dfrac{12\,R_E}{5.692} = -2.108\,R_E$.

#### Member BG (diagonal)
Same section; take moments about $O$, the point where $BC$ (slope $\tfrac13$) produced meets the bottom chord produced, at $x = -6$ m. ($AB$ is steeper, with slope $\tfrac23$, so it is not on this line.) Lever arm of $BG$ about $O$ $= 9.985$ m.

- Load right of $G$: $F_{BG} = +\dfrac{6\,R_A}{9.985} = +0.601\,R_A$ (tension).
- Load left of $H$: $F_{BG} = -\dfrac{30\,R_E}{9.985} = -3.005\,R_E$ (compression).

#### Ordinates (per unit load; + tension, − compression)
| Member | A | H | G | F | E |
|---|---|---|---|---|---|
| BC | 0.000 | -0.527 | -1.054 | -0.527 | 0.000 |
| BG | 0.000 | -0.751 | 0.300 | 0.150 | 0.000 |

Between $H$ and $G$ the ILDs are straight lines; $BG$ changes sign at $x = 6 + 6\times\dfrac{0.751}{0.751+0.300} = 10.3$ m from $A$.

#### Maximum force in BC for 6 kN/m over 8 m
All ordinates of $BC$ are negative, so the 8 m load is placed to enclose the largest area, symmetrically about the centre ($x = 8$ m to $16$ m):

Ordinates at $x=8$: $-0.703$; at $x=12$: $-1.054$; at $x=16$: $-0.703$.

$$
\text{area} = 2\times\frac{(0.703+1.054)}{2}\times 4 = 7.027\ \text{m}, \qquad F_{BC,max} = 6\times 7.027 = 42.16\ \text{kN}
$$

**Answer: maximum force in $BC$ = 42.16 kN (compression), with the 8 m load placed symmetrically about mid-span.**

## Check yourself

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

1. Which joint is used to find BC in the section through BC, BG and HG?
   A. Joint G, where BG and HG meet
   B. Joint B
   C. Joint A
   D. Point O at x = -6 m
   Answer: A. BG and HG meet at G, so moments about G give BC alone (lever arm 5.692 m). (Key point: Moment centre for BC)

2. What is the nature of the BC ILD?
   A. Tension with peak at G
   B. All ordinates negative (compression), peak -1.054 at G
   C. Changes sign at G
   D. Zero everywhere
   Answer: B. Ordinates are -0.527, -1.054, -0.527, all negative. (Key point: Sign of BC ILD)

3. How is the 6 kN/m, 8 m long load placed for maximum BC?
   A. At the left end
   B. Symmetrically about the centre, x = 8 to 16 m, to enclose the largest area
   C. At the right end
   D. Only over the panel HG
   Answer: B. All ordinates have the same sign, so the load is centred on the peak ordinate to give maximum area. (Key point: Placing the 8 m UDL)

4. What is the maximum force in BC?
   A. 7.03 kN
   B. 42.16 kN (compression)
   C. 50.6 kN (tension)
   D. 84.3 kN (compression)
   Answer: B. Area 7.027 m times 6 kN/m = 42.16 kN, compression. (Key point: Maximum BC force)

## 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 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
- [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: 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

## Related in other subjects

- [Truss indeterminacy; forces in BC, BG, HG, GD](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-indeterminacy-bc-bg-hg-gd) — Applied Mechanics, chapter 7
- [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
- [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 BG, HC, CF](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Atruss-bar-forces-bg-hc-cf) — Theory of Structures II, chapter 2