---
title: "Truss member forces BC, BD, BE, DE, DG, EG"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-bc-bd-be-de-dg-eg-48kn"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-bc-bd-be-de-dg-eg-48kn
site: BCT Engineering
---

# Truss member forces BC, BD, BE, DE, DG, EG

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2069 Chaitra, 7 marks
- Video: [Calculate the forces in members BC, BE, and EF. | Trusses Method of Sections |  Engineers Academy](https://www.youtube.com/watch?v=ixBSF5L7-sQ) — Engineers Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-bc-bd-be-de-dg-eg-48kn
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Compute the force developed in the member BC, BD, BE, DE, DG and EG of the given truss loaded as shown in figure. [Figure: truss with pin support at A and roller/pin support at G; top chord A-B-D-F-H with panels of 4 m each; lower joints C and E; 48 kN load at A; height at right 4.5 m.]

## Short answer

BC, BE, DE, and DG are zero-force members. BD = 128 kN (tension) and EG = 136.70 kN (compression).

## Answer

#### Data and assumptions
The top chord A–B–D–F–H is horizontal with 4 m panels: A(0,0), B(4,0), D(8,0), F(12,0), H(16,0). The lower chord runs in a straight line from A down to G, so C(4,−1.5), E(8,−3), G(12,−4.5) (depth 4.5 m under F). Members: AB, BD, DF, FH, AC, CE, EG (chords), BC, DE, FG (verticals), BE, DG (diagonals) and HG. The figure shows a vertical 48 kN load at A and supports at G and H. (m = 13, r = 3, j = 8: determinate.) Whichever of G and H is taken as the pin, the horizontal reaction is zero and the member forces are the same.

#### Reactions
- ΣM_H = 0: 48(16) = G_y(4), so **G_y = 192 kN ↑**; H_y = 144 kN ↓ (the support at H holds the truss down). The horizontal reactions are zero.

#### Method of joints
Direction of the lower chord: AC = CE = EG = √(4² + 1.5²) = 4.272 m, cos = 0.9363, sin = 0.3511.

**Joint A** (48 kN ↓): ΣFy: −48 − 0.3511F_AC = 0, so **F_AC = −136.70 kN (C)**. ΣFx: F_AB + 0.9363(−136.70) = 0, so **F_AB = +128.00 kN (T)**.

**Joint C**: AC and CE lie on one straight line and BC is the only other member, with no load at C, so **F_BC = 0** and **F_CE = F_AC = −136.70 kN (C)**.

**Joint B**: members AB, BD (collinear), BC = 0 and BE. Since BC = 0, ΣFy gives −0.6F_BE = 0, so **F_BE = 0** and F_BD = F_AB = **+128.00 kN (T)**.

**Joint E**: CE and EG are collinear and BE = 0, so **F_DE = 0** and **F_EG = F_CE = −136.70 kN (C)**.

**Joint D**: BD = 128 kN (T) and DE = 0, so DF = 128 kN (T), and ΣFy gives −0.7474F_DG = 0, so **F_DG = 0**.

**Joint G** (check): EG pushes G with (128 →, 48 ↓); HG (compression 192.67 kN) pushes G with (128 ←, 144 ↓); G_y = 192 kN ↑ balances 48 + 144 = 192 ✓.

#### Result
| Member | Force (kN) | Nature |
|---|---|---|
| BC | 0 | zero force |
| BD | 128.00 | tension |
| BE | 0 | zero force |
| DE | 0 | zero force |
| DG | 0 | zero force |
| EG | 136.70 | compression |

BC, BE, DE and DG are zero-force members: at each unloaded joint (C, B, E, D in turn), once the zero members already found are removed, the member is the only one not collinear with the others.

## Check yourself

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

1. Taking moments about H with the 48 kN load at A (16 m away) and G 4 m from H, what is G_y?
   A. 48 kN up
   B. 16 kN up
   C. 192 kN up
   D. 144 kN up
   Answer: C. 48(16) = G_y(4) gives G_y = 192 kN up; the support at H then holds the truss down with 144 kN. (Key point: Reaction G_y)

2. At joint A (48 kN down, AC sloping down at sin = 0.3511), what is F_AC?
   A. 48 kN tension
   B. 136.7 kN compression
   C. 136.7 kN tension
   D. 128 kN compression
   Answer: B. -48 - 0.3511F_AC = 0 gives F_AC = -136.7 kN, so AC is in compression. (Key point: Member AC from joint A)

3. Why is F_BC = 0 at joint C?
   A. BC is vertical and so cannot carry load
   B. AC and CE are collinear, no load acts at C, so the third member BC must be zero
   C. The load at A is far from C
   D. BC is in compression equal to AC
   Answer: B. At a loadless joint with two collinear members and one other member, the third member has no component to balance and carries zero force. (Key point: Zero-force member BC)

4. Once BC = 0, what follows at joint B and what is F_BD?
   A. F_BE = 0 and F_BD = F_AB = 128 kN (T)
   B. F_BE = 128 kN and F_BD = 0
   C. F_BE = 48 kN and F_BD = 80 kN
   D. Both BE and BD are zero
   Answer: A. With BC zero, ΣFy forces BE to be zero, so the collinear chord members AB and BD carry equal force of 128 kN tension. (Key point: Chain of zero forces)

## Related in Applied Mechanics

- [Truss forces in CE, BE, BD, DE; stability](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-ce-be-bd-de-stability-determinacy) — 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
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- [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
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## Related in other subjects

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