---
title: "Truss forces in CE, BE, BD, DE; stability"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-forces-ce-be-bd-de-stability-determinacy"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-forces-ce-be-bd-de-stability-determinacy
site: BCT Engineering
---

# Truss forces in CE, BE, BD, DE; stability

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2071 Shrawan, 5+3 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-forces-ce-be-bd-de-stability-determinacy
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Find the member forces in CE, BE, BD and DE for the given truss. Define stability and determinacy of structures with examples. [Figure: truss with pin support at A and roller at H, panel span 5@4 m = 20 m, heights 3 m and 5 m; loads 5 kN at B, 10 kN at 30° at D and 20 kN downward at E.]

## Short answer

Method of sections and joints: CE = 22.03 kN (T), BE = 1.33 kN (T), BD = 34.28 kN (C), DE = 17.00 kN (T). Stability means a structure maintains its shape under load without rigid-body motion; determinacy is when m + r = 2j (no redundancy).

## Answer

#### Data
A(0,0) pin and H(20,0) roller; bottom joints C(5,0), E(10,0), G(15,0); top joints B(5,3), D(10,5), F(15,3). The figure shows four equal panels between A and H (the label 5@4 m = 20 m gives the same 20 m span). Loads: 5 kN ↓ at B; 10 kN at D at 30° to the horizontal pointing down-left (8.66 kN ←, 5 kN ↓); 20 kN ↓ at E. (m = 13, r = 3, j = 8: determinate.)

#### Reactions
- ΣM_A = 0: 20H_y − 5(5) − 5(10) − 20(10) + 8.66(5) = 0, so **H_y = 11.58 kN ↑**.
- ΣFy: A_y = 30 − 11.58 = **18.42 kN ↑**; ΣFx: **A_x = 8.66 kN →**.

#### Section through BD, BE and CE (left part: A, B, C)
BD = √(5² + 2²) = 5.385 m (cos 0.9285, sin 0.3714); BE = √(5² + 3²) = 5.831 m (cos 0.8575, sin 0.5145).
- **CE**: ΣM_B = 0: 3F_CE − 18.42(5) + 8.66(3) = 0, so **F_CE = +22.03 kN (T)**.
- **BD**: ΣM_E = 0 (anticlockwise +): −18.42(10) + 5(5) − 4.643F_BD = 0, so **F_BD = −34.28 kN (C)**.
- **BE**: ΣFy: 18.42 − 5 + 0.3714(−34.28) − 0.5145F_BE = 0, so **F_BE = +1.33 kN (T)**.

#### Joint D (for DE)
BD (compression) pushes D with (31.83 →, 12.73 ↑); DF goes down-right (0.9285, −0.3714); DE is vertical.
- ΣFx: 31.83 + 0.9285F_DF − 8.66 = 0, so F_DF = −24.96 kN (C).
- ΣFy: 12.73 + 0.3714(24.96) − F_DE − 5 = 0, so **F_DE = +17.00 kN (T)**.

#### Result
| Member | Force (kN) | Nature |
|---|---|---|
| CE | 22.03 | tension |
| BE | 1.33 | tension |
| BD | 34.28 | compression |
| DE | 17.00 | tension |

#### Stability and determinacy
- **Stability:** a structure is stable if it keeps its shape and position under any load (supports prevent rigid-body motion and the members form a rigid arrangement of triangles). Example: the triangular truss above on a pin and a roller. A truss with a square panel without a diagonal, or supports whose reactions are all parallel or concurrent, is unstable.
- **Determinacy:** externally r − 3 and for a truss m + r − 2j (= 0 determinate, > 0 indeterminate, < 0 unstable). The truss above: 13 + 3 − 16 = 0, so it is determinate. If a diagonal is added to a panel already braced, m + r − 2j = 1 and the truss becomes indeterminate to the first degree.

## Check yourself

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

1. Which moment centre gives CE directly from the section?
   A. Joint E
   B. Joint A
   C. Joint B, where BD and BE intersect
   D. Joint H
   Answer: C. BD and BE both pass through B, so moments about B contain only CE. (Key point: Moment centre for CE)

2. What is the force in BD?
   A. 34.28 kN tension
   B. 34.28 kN compression
   C. 22.03 kN compression
   D. 17.00 kN tension
   Answer: B. ΣM_E gives a negative value, so BD = -34.28 kN, compression. (Key point: Force in BD)

3. How is DE = 17.00 kN tension found?
   A. From section 1-1 directly
   B. By assuming symmetry
   C. From ΣFy at joint D using the forces in BD and DF and the loads
   D. From moments about A
   Answer: C. At joint D, vertical equilibrium with BD's vertical push, DF's component and the 5 kN load gives F_DE = +17.00 kN. (Key point: DE from joint D)

4. If a diagonal is added to a panel that is already braced, m + r - 2j becomes:
   A. 0, determinate
   B. 1, indeterminate to first degree
   C. -1, unstable
   D. 3
   Answer: B. An extra member adds one to m, so m + r - 2j = 1 and the truss is indeterminate to the first degree. (Key point: Determinacy of truss)

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