---
title: "Truss member forces; assumptions of perfect truss"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-forces-assumptions-20kn-30kn-45deg"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-forces-assumptions-20kn-30kn-45deg
site: BCT Engineering
---

# Truss member forces; assumptions of perfect truss

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2073 Chaitra, 6+2 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-assumptions-20kn-30kn-45deg
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Determine the force developed in the members of the given truss. What are the assumptions of perfect truss? [Figure: truss with joints A, B, C, D, E; supports at A and C; 20 kN load at D; 30 kN load at E at 45°; top chord with dimensions 1 m, 4 m, 1 m; bottom 3 m + 3 m; height 2 m.]

## Short answer

Method of joints: AB = 13.64 kN (T), BC = 28.18 kN (T), AD = 30.49 kN (C), DE = 20.91 kN (C), EC = 15.59 kN (C), DB = 10.29 kN (T), BE = 10.29 kN (C). A perfect truss assumes frictionless pins, loads at joints, negligible weight, axial forces only, and rigid geometry.

## Answer

#### Data
A(0,0) roller, B(3,0) and C(6,0) pin on the bottom; D(1,2) and E(5,2) on the top (top chord 1 m + 4 m + 1 m, height 2 m). Members: AB, BC, AD, DE, EC, DB, BE (m = 7, j = 5, r = 3: determinate). Loads: 20 kN ↓ at D; 30 kN at E at 45° pointing down-left (21.21 kN ← and 21.21 kN ↓).
Lengths: AD = EC = √(1² + 2²) = 2.236 m (sin = 0.8944, cos = 0.4472); DB = BE = 2.828 m (45°).

#### Reactions
- ΣM_C = 0 (anticlockwise +): −6A_y + 20(5) + 21.21(1) + 21.21(2) = 0, so **A_y = 27.27 kN ↑**.
- ΣFx: **C_x = 21.21 kN →**. ΣFy: C_y = 20 + 21.21 − 27.27 = **13.94 kN ↑**.

#### Method of joints
**Joint A**: ΣFy: 27.27 + 0.8944F_AD = 0, so **F_AD = −30.49 kN (C)**. ΣFx: F_AB + 0.4472(−30.49) = 0, so **F_AB = +13.64 kN (T)**.

**Joint D**: AD pushes D with (13.64 →, 27.27 ↑).
- ΣFy: 27.27 − 20 − 0.7071F_DB = 0, so **F_DB = +10.29 kN (T)**.
- ΣFx: 13.64 + F_DE + 0.7071(10.29) = 0, so **F_DE = −20.91 kN (C)**.

**Joint B**: AB pulls B to the left (13.64 kN); DB goes up-left and BE up-right at 45°. No external load acts at B, so ΣFy: 0.7071(10.29) + 0.7071F_BE = 0 gives **F_BE = −10.29 kN (C)**. ΣFx: −13.64 − 7.27 − 7.27 + F_BC = 0, so **F_BC = +28.18 kN (T)**.

**Joint C**: ΣFy: 13.94 + 0.8944F_EC = 0, so **F_EC = −15.59 kN (C)**. Check ΣFx: 21.21 − 28.18 + 0.4472(15.59) = 0 ✓.

#### Result
| Member | Force (kN) | Nature |
|---|---|---|
| AB | 13.64 | tension |
| BC | 28.18 | tension |
| AD | 30.49 | compression |
| DE | 20.91 | compression |
| EC | 15.59 | compression |
| DB | 10.29 | tension |
| BE | 10.29 | compression |

#### Assumptions of a perfect truss
1. All members are straight and joined at their ends by smooth (frictionless) pins.
2. Loads and support reactions act only at the joints.
3. The self-weight of members is negligible (or is lumped at the joints).
4. Every member is a two-force member, so it carries only axial force (tension or compression), with no bending or shear.
5. Members are rigid enough that the truss geometry does not change appreciably under load, and all members lie in one plane.

## Check yourself

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

1. How is the 30 kN load at 45 degrees handled?
   A. Applied as 30 kN vertical
   B. Resolved into 21.21 kN horizontal and 21.21 kN vertical components
   C. Ignored in the reactions
   D. Split into 15 kN each
   Answer: B. 30 cos45 = 30 sin45 = 21.21 kN, used in both the moment and force equations. (Key point: Resolving the inclined load)

2. What is the force in AD?
   A. 30.49 kN tension
   B. 27.27 kN compression
   C. 13.64 kN tension
   D. 30.49 kN compression
   Answer: D. ΣFy at A: 27.27 + 0.8944F_AD = 0 gives F_AD = -30.49 kN, compression. (Key point: Joint A)

3. Why does F_BE = -F_DB (10.29 kN each, one tension and one compression)?
   A. Because AB equals BC
   B. Because B is a support
   C. Because no vertical external load acts at B, so the vertical components of DB and BE must cancel
   D. Because both are zero
   Answer: C. ΣFy at B: 0.7071F_DB + 0.7071F_BE = 0, so BE = -DB. (Key point: Joint B)

4. Which is NOT an assumption of a perfect truss?
   A. Members are straight
   B. Joints are smooth pins
   C. Members carry only axial force
   D. Members resist bending at the joints
   Answer: D. Perfect trusses assume frictionless pins, so there is no bending or shear in members. (Key point: Assumptions)

## Related in Applied Mechanics

- [Truss member forces; assumptions of perfect truss](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-member-forces-assumptions-15kn-10kn) — Applied Mechanics, chapter 7
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