---
title: "Truss indeterminacy; forces in BC, BG, HG, GD"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-indeterminacy-bc-bg-hg-gd"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-indeterminacy-bc-bg-hg-gd
site: BCT Engineering
---

# Truss indeterminacy; forces in BC, BG, HG, GD

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2070 Chaitra, 2+6 marks
- Video: [Understanding and Analysing Trusses](https://www.youtube.com/watch?v=Hn_iozUo9m4) — The Efficient Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-indeterminacy-bc-bg-hg-gd
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Determine the total degree of internal and external indeterminacy of the given truss. Also determine the member forces in members BC, BG, HG and GD. [Figure: truss with pin support at A and roller at E, bottom joints A, H, G, F, E at 4 m spacing (4 panels of 4 m), top joints B, C, D, height 4 m; loads 10 kN horizontal at D, 5 kN downward at H, 10 kN downward at G, 15 kN downward at F.]

## Short answer

The truss is statically determinate (external and internal indeterminacy zero). By method of joints, the required member forces are: BC = 15.00 kN compression, BG = 7.07 kN tension, HG = 20.00 kN tension, GD = 7.07 kN tension.

## Answer

#### Degree of indeterminacy
Joints: j = 8 (A, B, C, D, E, F, G, H). Members: AB, BC, CD, DE (top chord and end rafters), AH, HG, GF, FE (bottom chord), BH, CG, DF (verticals), BG, GD (diagonals): m = 13. Reactions: pin at A (2) + roller at E (1): r = 3.
- External indeterminacy: D_e = r − 3 = 3 − 3 = **0**.
- Internal indeterminacy: D_i = m − (2j − 3) = 13 − 13 = **0**.
- Total: m + r − 2j = 13 + 3 − 16 = 0. The truss is statically determinate and stable (all panels triangulated).

#### Data
A(0,0), H(4,0), G(8,0), F(12,0), E(16,0) on the bottom (4 m panels); B(4,4), C(8,4), D(12,4) on the top (height 4 m). Loads: 10 kN → at D, 5 kN ↓ at H, 10 kN ↓ at G, 15 kN ↓ at F.

#### Reactions
- ΣM_A = 0: 16E_y − 5(4) − 10(8) − 15(12) − 10(4) = 0, so **E_y = 20 kN ↑**.
- ΣFy: A_y = 30 − 20 = **10 kN ↑**; ΣFx: **A_x = 10 kN ←**.

#### Method of joints
**Joint A** (AB at 45°): ΣFy: 10 + 0.7071F_AB = 0, so F_AB = −14.14 kN (C). ΣFx: −10 + 0.7071(−14.14) + F_AH = 0, so F_AH = +20 kN (T).

**Joint H**: ΣFy: −5 + F_BH = 0, so F_BH = +5 kN (T). ΣFx: −20 + F_HG = 0, so **F_HG = +20.00 kN (T)**.

**Joint B** (AB pushes B with 10 kN → and 10 kN ↑; BH pulls down 5 kN):
- ΣFy: 10 − 5 − 0.7071F_BG = 0, so **F_BG = +7.07 kN (T)**.
- ΣFx: 10 + F_BC + 0.7071(7.07) = 0, so **F_BC = −15.00 kN (C)**.

**Joint C**: BC and CD are collinear and no load acts at C, so F_CG = 0 and F_CD = −15 kN (C).

**Joint G** (HG = 20 T pulls left, BG pulls up-left, GD goes up-right):
- ΣFy: −10 + 0.7071(7.07) + 0.7071F_GD + F_CG(0) = 0, so **F_GD = +7.07 kN (T)**.
- ΣFx: −20 + F_GF − 0.7071(7.07) + 0.7071(7.07) = 0, so F_GF = +20 kN (T).

#### Result
| Member | Force (kN) | Nature |
|---|---|---|
| BC | 15.00 | compression |
| BG | 7.07 | tension |
| HG | 20.00 | tension |
| GD | 7.07 | tension |

Remaining members (check at D and E): DF = +15 T, DE = −28.28 C, EF = +20 T, CD = −15 C.

## Check yourself

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

1. The truss has j = 8, m = 13 and r = 3. What are the external, internal and total indeterminacy?
   A. External 1, internal 0, total 1
   B. External 0, internal 0, total 0
   C. External 0, internal 1, total 1
   D. External 3, internal 13, total 16
   Answer: B. External = r - 3 = 0, internal = m - (2j - 3) = 13 - 13 = 0, total = m + r - 2j = 0, so it is statically determinate. (Key point: Degree of indeterminacy)

2. Which moment equation gives E_y for this truss (pin A, roller E, 16 m span)?
   A. Moments about A with only the vertical loads: 16E_y = 30 x 8
   B. Moments about E ignoring the 10 kN horizontal load
   C. Moments about A including the 10 kN horizontal load at height 4 m: 16E_y = 5(4)+10(8)+15(12)+10(4), E_y = 20 kN
   D. Sum of vertical forces: E_y = 30 kN
   Answer: C. The horizontal 10 kN load at D has a moment about A (10 x 4), which must be included, giving E_y = 320/16 = 20 kN. (Key point: Reaction at roller E)

3. At joint H there is a 5 kN downward load, vertical member BH and bottom chord members AH and HG. How is F_HG obtained?
   A. F_BH = 5 kN (T) from ΣFy, then ΣFx gives F_HG = F_AH = 20 kN (T)
   B. F_HG = 5 kN because it carries the load
   C. F_HG = 0 since H has only a vertical load
   D. F_HG = 10 kN from the horizontal load at D
   Answer: A. At H, BH balances the vertical load; the chord members are collinear so F_HG equals F_AH = 20 kN tension. (Key point: Force in HG at joint H)

4. At joint B, which sign and size results for F_BC?
   A. 15 kN tension
   B. 7.07 kN compression
   C. 15 kN compression
   D. 20 kN tension
   Answer: C. Using ΣFx at B with AB's horizontal push of 10 kN and BG's tension component 5 kN: 10 + F_BC + 5 = 0, so F_BC = -15 kN (compression). (Key point: Force in BC)

## Related in Applied Mechanics

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- [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
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## Related in other subjects

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- [Truss ILDs for BC and BG; max force in BC](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-bc-bg-live-load-8m) — Theory of Structures I, chapter 5
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