---
title: "Truss member forces with 25, 50, 10 kN loads"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-forces-25kn-50kn-10kn-hinge-roller"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-forces-25kn-50kn-10kn-hinge-roller
site: BCT Engineering
---

# Truss member forces with 25, 50, 10 kN loads

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2070 Asar, 8 marks
- Video: [Statics-6.2 -  Determine the force in each member of the truss.](https://www.youtube.com/watch?v=_WjsuWaABDU) — Learning by Teaching
- 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-25kn-50kn-10kn-hinge-roller
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Determine the member forces for given truss loaded as shown in figure below. [Figure: truss with joints A, B, C on top (spans 6 m and 6 m), D and E at bottom (3 m, 6 m, 3 m); loads 25 kN at A, 50 kN at B and 10 kN at C, downward; a hinge support at E (hinge) and roller at E side, depth 3 m.]

## Short answer

Reactions: 200 kN ↑ at E, 115 kN ↓ at C, no horizontal. Using method of joints: AB 25 kN T, BC 125 kN T, AD 35.36 kN C, DB 35.36 kN T, BE 106.07 kN C, EC 176.78 kN C, DE 50 kN C.

## Answer

#### Data
Top joints A(0,3), B(6,3), C(12,3); bottom joints D(3,0), E(9,0). Members: AB, BC, AD, DB, BE, EC, DE (m = 7, j = 5, r = 3: determinate). Loads: 25 kN at A, 50 kN at B and 10 kN at C, all downward. Support: hinge (pin) at C and roller at E (vertical reaction). All sloping members are at 45° (3 m rise over 3 m run, length 4.243 m).

#### Reactions
- ΣM_C = 0: 25(12) + 50(6) − E_y(3) = 0, so **E_y = 200 kN ↑**.
- ΣFy: C_y = 85 − 200 = **−115 kN (115 kN ↓)**; ΣFx: **C_x = 0**.

#### Method of joints
**Joint A**: ΣFy: −25 − 0.7071F_AD = 0, so **F_AD = −35.36 kN (C)**. ΣFx: F_AB + 0.7071(−35.36) = 0, so **F_AB = +25.00 kN (T)**.

**Joint D**: AD is in compression and pushes D with 0.7071 × 35.36 = 25 kN to the right and 25 kN downward.
- ΣFy: −25 + 0.7071F_DB = 0, so **F_DB = +35.36 kN (T)**.
- ΣFx: 25 + 0.7071(35.36) + F_DE = 0, so **F_DE = −50.00 kN (C)**.

**Joint B** (AB = 25 T pulls left, DB = 35.36 T pulls down-left):
- ΣFy: −50 − 0.7071(35.36) − 0.7071F_BE = 0, so **F_BE = −106.07 kN (C)**.
- ΣFx: −25 − 25 + 0.7071(−106.07) + F_BC = 0, so **F_BC = +125.00 kN (T)**.

**Joint E**: BE is in compression and pushes E with 75 kN to the right and 75 kN downward; DE (compression) pushes E with 50 kN to the right. ΣFy: 200 − 75 + 0.7071F_EC = 0, so **F_EC = −176.78 kN (C)**. Check ΣFx: 50 + 75 − 0.7071(176.78) = 0 ✓.

#### Result
| Member | Force (kN) | Nature |
|---|---|---|
| AB | 25.00 | tension |
| BC | 125.00 | tension |
| AD | 35.36 | compression |
| DB | 35.36 | tension |
| BE | 106.07 | compression |
| EC | 176.78 | compression |
| DE | 50.00 | compression |

## Check yourself

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

1. For the truss with 25, 50 and 10 kN downward loads, which equation gives the roller reaction E_y?
   A. Moments about C: 25(12) + 50(6) - E_y(3) = 0, so E_y = 200 kN up
   B. Sum of vertical forces only: E_y = 25 + 50 + 10 = 85 kN up
   C. Moments about E: 25(12) + 50(6) = 10(3), so E_y = 100 kN
   D. Moments about C with the 10 kN load included at 6 m, giving 125 kN
   Answer: A. Taking moments about the hinge C removes C_x and C_y, leaving E_y as the only unknown: E_y = 600/3 = 200 kN upward. (Key point: Support reaction by moments)

2. From the vertical equilibrium, C_y = 85 - 200 = -115 kN. What does this mean?
   A. The truss is unstable
   B. The hinge pushes up with 115 kN
   C. The hinge reaction is 115 kN acting downward (holding the truss down)
   D. The calculation must be repeated with the opposite sign convention for loads
   Answer: C. The minus sign means the assumed upward direction is wrong; the hinge pulls the truss down with 115 kN because the roller reaction exceeds the total load. (Key point: Meaning of negative C_y)

3. At joint A the only vertical forces are the 25 kN downward load and the vertical component of AD. What is F_AD?
   A. 35.36 kN tension
   B. 25 kN compression
   C. 25 kN tension
   D. 35.36 kN compression
   Answer: D. AD must push A upward to balance the load, so it is in compression: F_AD = 25/0.7071 = 35.36 kN (C). (Key point: Joint A: nature of AD)

4. At joint D, AD (compression) pushes D with 25 kN to the right and 25 kN down. How is F_DE found and what is it?
   A. From ΣFx with DB's horizontal part: DE = 50 kN compression
   B. From ΣFx with DB's horizontal part: DE = 50 kN tension
   C. From ΣFy alone: DE = 25 kN tension
   D. From ΣFx ignoring DB: DE = 25 kN compression
   Answer: A. DB is tension 35.36 kN (pulls D up and right giving 25 kN horizontal), so 25 + 25 + F_DE = 0 gives F_DE = -50 kN, i.e. compression. (Key point: Joint D: force in DE)

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