---
title: "Truss ILDs for bc, hg, df, upper chord"
subject: "Theory of Structures I"
chapter: 5
key: "semester_4/theory_of_structures_i:truss-ild-bc-hg-df-upper-chord"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/truss-ild-bc-hg-df-upper-chord
site: BCT Engineering
---

# Truss ILDs for bc, hg, df, upper chord

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

- Asked in: 2075 Baisakh, 10 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-hg-df-upper-chord
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-5

## Question

Draw influence line diagrams for the forces in member bc, hg and df of the truss. The load moves in the upper chord of the truss. [Figure: truss with bottom joints a (hinge), b, c, d, e (roller), 4 panels of 10 m = 40 m; top joints j, i, h, g, f at a height of 12 m; members as drawn.]

## Short answer

For load on upper chord: member bc ILD is tension, peak +0.625 at i, zero at j and f. Member hg ILD is compression, peak -0.833 at h, -0.417 at i and g. Member df ILD is tension, rising from 0 at j to +0.976 at g, then 0 at f.

## Answer

**Assumed truss (Pratt type, figure not available):** bottom joints $a, b, c, d, e$ and top joints $j, i, h, g, f$ directly above them, 4 panels of 10 m (span 40 m), height 12 m; verticals $aj, ib, hc, gd, fe$; diagonals $jb, ic, gc, fd$ (sloping down towards the centre). $a$ hinge, $e$ roller. The load moves on the **upper (deck) chord**, so each ILD is straight between the top joints.

Reactions for a unit load at $x$ from $a$: $R_a = \dfrac{40-x}{40}$, $R_e = \dfrac{x}{40}$. Diagonal $df$ has length $\sqrt{10^2+12^2} = 15.62$ m, $\sin\theta = 12/15.62 = 0.768$.

#### Method of sections
**Member bc (bottom chord, panel $bc$):** cut $ih$, $ic$, $bc$; take moments about $i$ (where $ih$ and $ic$ meet), lever arm 12 m.

- Load right of $h$: $F_{bc} = \dfrac{R_a\times 10}{12} = 0.833R_a$ (tension).
- Load left of $i$: $F_{bc} = \dfrac{R_e\times 30}{12} = 2.5R_e$.

**Member hg (top chord, panel $cd$):** cut $hg$, $gc$, $cd$; moments about $c$, lever arm 12 m.

- Load right of $g$: $F_{hg} = -\dfrac{R_a\times 20}{12} = -1.667R_a$ (compression).
- Load left of $h$: $F_{hg} = -\dfrac{R_e\times 20}{12} = -1.667R_e$.

**Member df (diagonal, panel $de$):** cut $gf$, $fd$, $de$; the panel shear is carried by the diagonal.

- Load left of $g$: $F_{df}\sin\theta = R_e \Rightarrow F_{df} = 1.302R_e$ (tension).
- Load at $f$ (over the support): 0. Between $g$ and $f$ the ILD is the straight line joining the ordinates.

#### Ordinates (per unit load; + tension, − compression)
| Member | j | i | h | g | f |
|---|---|---|---|---|---|
| bc | 0.000 | 0.625 | 0.417 | 0.208 | 0.000 |
| hg | 0.000 | -0.417 | -0.833 | -0.417 | 0.000 |
| df | 0.000 | 0.325 | 0.651 | 0.976 | 0.000 |

#### Shapes
- **bc:** tension, peak $+0.625$ at $i$, falling to $+0.417$ at $h$, $+0.208$ at $g$, 0 at $j$ and $f$.
- **hg:** compression, peak $-0.833$ under $h$, $-0.417$ at $i$ and at $g$.
- **df:** tension, rising from 0 at $j$ to $+0.976$ at $g$, then falling to 0 at $f$.

## Check yourself

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

1. What does 'load moves on the upper chord' imply for the ILD shape?
   A. The ILD is straight between the top joints
   B. The ILD is curved
   C. The ILD is constant
   D. Only the bottom joints matter
   Answer: A. Loads transfer at the top joints, so each ILD is linear between top joint ordinates. (Key point: Upper chord loading)

2. To find F_bc, moments are taken about which joint, and what is the lever arm?
   A. Joint c, 12 m
   B. Joint i, where ih and ic meet, 12 m
   C. Joint h, 10 m
   D. Joint b, 12 m
   Answer: B. Cutting ih, ic and bc, the two unknowns ih and ic meet at i, so moments about i give F_bc with a 12 m lever arm. (Key point: Moment centre for bc)

3. Which describes the ILD of top-chord member hg?
   A. Tension with peak +0.833 at h
   B. Compression with peak -0.833 under h
   C. Compression with peak -1.667 at j
   D. Zero everywhere
   Answer: B. hg is a top chord in compression: 1.667 R = 1.667 x 0.5 = 0.833 at h, with -0.417 at i and g. (Key point: Sign and peak of hg)

4. How is the force in diagonal df related to the panel shear?
   A. F sin(theta) = R_e, so F = 1.302 R_e (load left of g)
   B. F = R_e / cos(theta)
   C. F = R_e x 12
   D. F = 0 for every load position
   Answer: A. The vertical component of df balances the shear R_e, with sin(theta) = 0.768 giving 1.302 R_e (tension). (Key point: Diagonal df)

## 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 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 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 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
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- [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 other subjects

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