---
title: "Member forces in given truss (2074 Asoj)"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-member-forces-15kn-5kn"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-member-forces-15kn-5kn
site: BCT Engineering
---

# Member forces in given truss (2074 Asoj)

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2074 Asoj, 8 marks
- Video: [Trusses Method of Joints | Mechanics Statics | Learn to Solve Questions](https://www.youtube.com/watch?v=_rK02neOF18) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-member-forces-15kn-5kn
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Find the member force in the indicated members of the truss shown below. [Figure: truss with span 4@5 m = 20 m, heights 3 m and 5 m; loads 5 kN at 30° at D, 15 kN at F, 10 kN at G; supports at A and H; the indicated members are marked with ticks.]

## Short answer

Reactions found from equilibrium; method of sections and joint analysis yield: BD = 13.82 kN (C), DF = 18.49 kN (C), DE = 9.50 kN (T), BE = 2.49 kN (T), CE = 15.03 kN (T).

## Answer

#### Data and assumptions
Joints: A(0,0), C(5,0), E(10,0), G(15,0), H(20,0) on the bottom chord; B(5,3), D(10,5), F(15,3) on the top. The four panels of 5 m make 20 m. A is a pin and H a roller. Loads: 5 kN at D at 30° to the horizontal, pointing down-right (4.33 kN → and 2.50 kN ↓), 15 kN ↓ at F and 10 kN ↓ at G. The members marked in the figure are taken as BD, DF, DE, BE and CE. (m = 13, r = 3, j = 8: determinate.)

#### Reactions
- ΣM_A = 0: 20H_y − 10(15) − 15(15) − 2.5(10) − 4.33(5) = 0, so 20H_y = 421.65 and **H_y = 21.08 kN ↑**.
- ΣFy: A_y = 27.5 − 21.08 = **6.42 kN ↑**; ΣFx: **A_x = 4.33 kN ←**.

#### Section 1–1 between B–C and D–E (cuts BD, BE, CE; left part)
Lengths: BD = √(5² + 2²) = 5.385 m, BE = √(5² + 3²) = 5.831 m.
- **CE**: ΣM_B = 0: 3F_CE − 6.42(5) − 4.33(3) = 0 → **F_CE = +15.03 kN (T)**.
- **BD**: ΣM_E = 0 (clockwise +): 6.42(10) + F_BD(0.3714×5 + 0.9285×3) = 0 → 64.17 + 4.643F_BD = 0 → **F_BD = −13.82 kN (C)**.
- **BE**: ΣFy: 6.42 + 0.3714F_BD − 0.5145F_BE = 0 → 6.42 − 5.13 = 0.5145F_BE → **F_BE = +2.49 kN (T)**.

#### Joint D (cos = 5/5.385 = 0.9285, sin = 2/5.385 = 0.3714)
- BD is in compression and pushes D with 13.82 × 0.9285 = 12.84 kN to the right and 13.82 × 0.3714 = 5.13 kN upward.
- ΣFx: 12.84 + 0.9285F_DF + 4.33 = 0 → **F_DF = −18.49 kN (C)**.
- ΣFy: 5.13 − 0.3714F_DF − F_DE − 2.5 = 0 → 5.13 + 6.87 − 2.5 = F_DE → **F_DE = +9.50 kN (T)**.

#### Result (indicated members)
| Member | Force (kN) | Nature |
|---|---|---|
| BD | 13.82 | compression |
| DF | 18.49 | compression |
| DE | 9.50 | tension |
| BE | 2.49 | tension |
| CE | 15.03 | tension |

Other members (for completeness): AB −12.47 C, FH −40.98 C, AC = CE = 15.03 T, EG = GH = 35.14 T, EF −20.96 C, FG = 10.00 T (it carries the load at G), BC = 0.

## Check yourself

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

1. Why is the moment about B used to find CE?
   A. B is a support
   B. CE is zero
   C. The other two cut members, BD and BE, pass through B and so have no moment about B
   D. It avoids using the reactions
   Answer: C. Taking moments about the intersection of two unknowns eliminates them, so only F_CE appears. (Key point: Choice of section)

2. What is the nature of BD?
   A. Compression, 19.21 kN
   B. Tension, 19.21 kN
   C. Compression, 23.87 kN
   D. Zero force
   Answer: A. The moment about E gives a negative F_BD, so BD is in compression with 19.21 kN. (Key point: Force in BD)

3. What is the force in CE?
   A. 19.19 kN compression
   B. 13.50 kN tension
   C. 19.19 kN tension
   D. Zero
   Answer: C. ΣM_B = 0 gives 3F_CE = 8.92(5) + 4.33(3), so F_CE = +19.19 kN (tension). (Key point: Force in CE)

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