---
title: "Truss member forces; assumptions of perfect truss"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-member-forces-assumptions-15kn-10kn"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-member-forces-assumptions-15kn-10kn
site: BCT Engineering
---

# Truss member forces; assumptions of perfect truss

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2074 Chaitra, 5+4 marks
- Video: [Assumptions and Determinacy for Truss Analysis - Statics and Structural Analysis](https://www.youtube.com/watch?v=7ZTJ13MW34w) — structurefree
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-member-forces-assumptions-15kn-10kn
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Find the member forces of the indicated members of the truss shown in the figure below. Write down the assumptions of a perfect truss. [Figure: truss with pin support at A, roller at B, joints F, E, D, C, G, H; loads 15 kN at E and 10 kN at C; span 6 ft @ 4 = 24 ft, height 7.5 ft.]

## Short answer

Member forces: BE = 19.21 kN (C), EG = 0, ED = 24.00 kN (T), DH = 0, GH = 26.00 kN (C). A perfect truss assumes straight members with frictionless pinned joints, loads only at joints, negligible self-weight, members carry only axial forces, and rigid geometry.

## Answer

#### Data
Take A(0,0) pin and B(6,0) roller on the bottom; F(6,7.5), E(12,7.5), D(18,7.5), C(24,7.5) on the top (6 ft panels, 7.5 ft high); the lower chord runs in a straight line from B to C, so G(12,2.5) and H(18,5). Loads: 15 kN ↓ at E, 10 kN ↓ at C. The members marked in the figure are taken as BE, EG, ED, DH and GH. (m = 13, r = 3, j = 8: determinate.)

#### Reactions
- ΣM_A = 0: 6B_y − 15(12) − 10(24) = 0, so **B_y = 70 kN ↑**.
- ΣFy: A_y = 25 − 70 = **−45 kN (45 kN ↓)**; ΣFx: A_x = 0.

#### Joint C (cut at the right-hand end)
Member CH has direction (−6, −2.5)/6.5 from C, so its vertical component is 2.5/6.5 = 0.3846.
- ΣFy: −10 − 0.3846F_CH = 0, so **F_HC = −26.0 kN (C)**.
- ΣFx: −0.9231F_CH − F_CD = 0, so F_CD = +24.0 kN (T).

#### Joint H
Members HG and HC lie on the same straight line and DH is the only other member with no load at H, so **F_DH = 0** and **F_GH = F_HC = −26.0 kN (C)**.

#### Joint D
ED and DC are collinear and equal (24 kN T); with DH = 0, DG must also be zero: **F_ED = +24.0 kN (T)**, F_DG = 0.

#### Joint G
BG and GH are collinear (26 kN C each) and DG = 0, so the vertical member **F_EG = 0**.

#### Joint E (cos = 6/9.605 = 0.6247, sin = 0.7809 for EB)
- ΣFy: −15 − 0.7809F_BE = 0 (E→B points down-left, EG = 0), so **F_BE = −19.21 kN (C)** (it pushes E upward with 15 kN).
- ΣFx: −F_FE + 24 + 0.6247(19.21) = 0 → F_FE = +36 kN (T) (check by section: ΣM_B on the left part: 7.5F_FE = 6×45 = 270, so F_FE = 36 kN).

#### Result (indicated members)
| Member | Force (kN) | Nature |
|---|---|---|
| BE | 19.21 | compression |
| EG | 0 | zero force |
| ED | 24.00 | tension |
| DH | 0 | zero force |
| GH | 26.00 | 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. Why is F_DH = 0 at joint H?
   A. H is a support
   B. DH is vertical
   C. HG and HC are collinear and no load acts at H, so the third member DH carries nothing
   D. The load at C is zero
   Answer: C. At an unloaded joint with two collinear members, the third member must be zero to balance forces perpendicular to the line. (Key point: Zero-force member DH)

2. How is By = 70 kN found?
   A. 6By = 15(12) + 10(24), moments about A
   B. By = 15 + 10
   C. By = 15 x 24
   D. ΣFx = 0
   Answer: A. Moments about the pin A: 6By = 180 + 240 = 420, so By = 70 kN. (Key point: Reaction at B)

3. What is the direction of Ay?
   A. 25 kN upward
   B. 70 kN upward
   C. Zero
   D. 45 kN downward
   Answer: D. ΣFy: Ay = 25 - 70 = -45, so the support at A pulls down with 45 kN. (Key point: Reaction at A)

4. Which is an assumption of a perfect truss?
   A. Loads act anywhere along members
   B. Joints are rigid
   C. Loads and reactions act only at the joints
   D. Members carry bending
   Answer: C. A perfect truss has smooth pinned joints and loads only at joints, so members carry axial force only. (Key point: Assumptions)

## Related in Applied Mechanics

- [Member forces of truss with 20 and 10 kN loads](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-20kn-10kn-12m-8m) — Applied Mechanics, chapter 7
- [Forces in all members of truss with 10.8 kN loads](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-10-8kn-d-support) — Applied Mechanics, chapter 7
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- [Member forces in truss with 20, 10, 12, 50 kN](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-20kn-10kn-12kn-50kn) — Applied Mechanics, chapter 7
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