---
title: "Forces in all members of truss with 10.8 kN loads"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-forces-10-8kn-d-support"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-forces-10-8kn-d-support
site: BCT Engineering
---

# Forces in all members of truss with 10.8 kN loads

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2068 Shrawan, 11 marks
- Video: [Truss Analysis using Joint Method 2](https://www.youtube.com/watch?v=BHNhJ65Ne8A) — Engineeringfun
- 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-10-8kn-d-support
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Determine the forces in all members of the truss shown in figure below. [Figure: truss with pin support at A and roller at C on the top chord, apex joint D below, B on the chord; 10.8 kN loads at B and at the section next to C; horizontal dimensions 22.5 m and 55[?] m; depth 12 m.]

## Short answer

Reactions: C_y = 15.03 kN ↑, A_y = 6.57 kN ↑, A_x = 0. By method of joints: AB 12.33 kN (C), BC 12.33 kN (C), BD 10.80 kN (C), CD 13.03 kN (T), AD 13.97 kN (T).

## Answer

#### Data and assumptions
A(0,0), B(22.5,0) and C(57.5,0) lie on the top chord (panels of 22.5 m and 35 m); D is 12 m below B. A is a pin and C a roller (vertical reaction); D is a free joint. Members: AB, BC, BD, CD and AD (m = 5, j = 4, r = 3, so m + r = 2j: determinate). Lengths: AB = 22.5 m, BC = 35 m, BD = 12 m, CD = √(35² + 12²) = 37 m (sin = 12/37 = 0.3243, cos = 35/37 = 0.9459), AD = √(22.5² + 12²) = 25.5 m (sin = 12/25.5 = 0.4706, cos = 22.5/25.5 = 0.8824). Loads: 10.8 kN ↓ at B and 10.8 kN ↓ at C.

```text
 A(pin)      B              C(roller)
  o---------o--------------o
    \       |          .
      \     |      .
        \   |  .
            D
```

#### Reactions
- ΣM_A = 0: 57.5C_y = 10.8(22.5) + 10.8(57.5) = 864, so **C_y = 15.03 kN ↑**.
- ΣFy: A_y = 21.6 − 15.03 = **6.57 kN ↑**; ΣFx: **A_x = 0**.

#### Method of joints (tension +, compression −)
**Joint A** (AB along +x; AD down-right):
- ΣFy: 6.57 − 0.4706F_AD = 0, so **F_AD = +13.97 kN (T)**.
- ΣFx: F_AB + 0.8824(13.97) = 0, so **F_AB = −12.33 kN (C)**.

**Joint B** (AB and BC collinear, BD vertical, 10.8 kN ↓):
- ΣFy: −10.8 − F_BD = 0, so **F_BD = −10.80 kN (C)**.
- ΣFx: F_BC = F_AB, so **F_BC = −12.33 kN (C)**.

**Joint C** (reaction 15.03 kN ↑, load 10.8 kN ↓; CD goes down-left):
- ΣFy: 15.03 − 10.8 − 0.3243F_CD = 0, so **F_CD = +13.03 kN (T)**.
- Check ΣFx: 12.33 − 0.9459(13.03) = 0 ✓.

**Joint D** (check): AD pulls D with (12.33 ←, 6.57 ↑), CD pulls D with (12.33 →, 4.23 ↑), and BD (compression) pushes D down with 10.80 kN: 6.57 + 4.23 − 10.80 = 0 ✓.

#### Result
| Member | Force (kN) | Nature |
|---|---|---|
| AB | 12.33 | compression |
| BC | 12.33 | compression |
| BD | 10.80 | compression |
| CD | 13.03 | tension |
| AD | 13.97 | tension |

The top chord and the vertical BD are in compression, and the lower members AD and CD are in tension. The 10.8 kN load at C acts directly over the roller, so it only adds to C_y and does not change any member force.

## Check yourself

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

1. With 10.8 kN down at C and CD sloping with sin = 0.3243, what is F_CD?
   A. 10.8 kN tension
   B. 33.30 kN tension
   C. 33.30 kN compression
   D. 31.50 kN compression
   Answer: C. CD must push C upward: 0.3243|F| = 10.8, so |F| = 33.30 kN, in compression. (Key point: Force in CD)

2. CD (compression) pushes C to the right with 0.9459 x 33.30 = 31.50 kN. What does BC do?
   A. BC is in compression 31.50 kN
   B. BC is in tension 31.50 kN, pulling C back
   C. BC is zero
   D. BC is in tension 33.30 kN
   Answer: B. To balance the rightward push, BC must pull C to the left, i.e. tension of 31.50 kN. (Key point: Force in BC)

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