---
title: "Member forces of truss with 20 and 10 kN loads"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-forces-20kn-10kn-12m-8m"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-forces-20kn-10kn-12m-8m
site: BCT Engineering
---

# Member forces of truss with 20 and 10 kN loads

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2067 Asar, 10 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-20kn-10kn-12m-8m
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Calculate the member forces of the given truss shown in figure below. [Figure: truss with joints A, B, C on top, D and E at bottom; 20 kN at A and 10 kN at B; top spans 12 m + 12 m; bottom spans 6 m, 12 m, 6 m; height 8 m; pin support at C and roller at E.]

## Short answer

Reactions: 100 kN ↑ at roller E, 70 kN ↓ at pin C. By method of joints: AB 15 kN T, BC 52.50 kN T, AD 25 kN C, DB 25 kN T, BE 37.50 kN C, EC 87.50 kN C, DE 30 kN C.

## Answer

#### Data
Top joints A(0,8), B(12,8), C(24,8); bottom joints D(6,0), E(18,0). Members: AB, BC, AD, DB, BE, EC, DE (m = 7, j = 5, r = 3: determinate). Loads: 20 kN ↓ at A and 10 kN ↓ at B. Supports: pin at C and roller at E (vertical reaction). The inclined members have length √(6² + 8²) = 10 m, so sin = 0.8 and cos = 0.6.

#### Reactions
- ΣM_C = 0 (anticlockwise +): 20(24) + 10(12) − 6E_y = 0, so **E_y = 100 kN ↑**.
- ΣFy: C_y = 30 − 100 = **−70 kN (70 kN ↓)**; ΣFx: **C_x = 0**.

#### Method of joints
**Joint A**: ΣFy: −20 − 0.8F_AD = 0, so **F_AD = −25.00 kN (C)**. ΣFx: F_AB + 0.6(−25) = 0, so **F_AB = +15.00 kN (T)**.

**Joint D** (AD compression pushes D with 15 kN → and 20 kN ↓):
- ΣFy: −20 + 0.8F_DB = 0, so **F_DB = +25.00 kN (T)**.
- ΣFx: 15 + 0.6(25) + F_DE = 0, so **F_DE = −30.00 kN (C)**.

**Joint B** (AB = 15 T pulls left, DB = 25 T pulls down-left; BE goes down-right):
- ΣFy: −10 − 0.8(25) − 0.8F_BE = 0, so **F_BE = −37.50 kN (C)**.
- ΣFx: −15 − 0.6(25) + 0.6(−37.50) + F_BC = 0, so **F_BC = +52.50 kN (T)**.

**Joint E**: BE pushes E with (22.5 →, 30 ↓) and DE (compression) pushes with 30 kN →. ΣFy: 100 − 30 + 0.8F_EC = 0, so **F_EC = −87.50 kN (C)**. Check ΣFx: 22.5 + 30 + 0.6(−87.5) = 0 ✓.

#### Result
| Member | Force (kN) | Nature |
|---|---|---|
| AB | 15.00 | tension |
| BC | 52.50 | tension |
| AD | 25.00 | compression |
| DB | 25.00 | tension |
| BE | 37.50 | compression |
| EC | 87.50 | compression |
| DE | 30.00 | compression |

## Check yourself

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

1. Taking moments about C, which equation gives E_y?
   A. 20(12) + 10(24) = 6E_y
   B. 20(24) + 10(12) = 6E_y, so E_y = 100 kN up
   C. 20(24) + 10(12) = 12E_y, so E_y = 50 kN
   D. E_y = 30 kN
   Answer: B. A is 24 m and B 12 m from C, and E is 6 m from C: E_y = (480 + 120)/6 = 100 kN up. (Key point: Roller reaction E_y)

2. At A with 20 kN down and AD of slope sin = 0.8, what is F_AD?
   A. 25 kN tension
   B. 16 kN compression
   C. 25 kN compression
   D. 20 kN compression
   Answer: C. AD pushes A upward: 20 = 0.8|F_AD|, so 25 kN compression. (Key point: Joint A: AD)

3. At D, DB (tension 25 kN) pulls up and right; AD pushes with 15 kN right. What is F_DE?
   A. 30 kN tension
   B. 15 kN compression
   C. 30 kN compression
   D. 10 kN compression
   Answer: C. ΣFx: 15 + 0.6(25) + F_DE = 0 gives F_DE = -30 kN, compression. (Key point: Joint D: DE)

4. At E, E_y = 100 kN up, BE pushes down 30 kN. What is F_EC?
   A. 87.5 kN tension
   B. 87.5 kN compression
   C. 70 kN compression
   D. 37.5 kN compression
   Answer: B. 100 - 30 + 0.8F_EC = 0 gives F_EC = -87.5 kN, compression. (Key point: Joint E: EC)

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