---
title: "Static determinacy and stability of a truss"
subject: "Strength of Materials"
chapter: 1
key: "semester_3/strength_of_materials:static-determinacy-stability-truss"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/static-determinacy-stability-truss
site: BCT Engineering
---

# Static determinacy and stability of a truss

Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments

- Asked in: 2067 Asar (old course), 4 marks
- Video: [Determinate vs Indeterminate Structures - Intro to Structural Analysis](https://www.youtube.com/watch?v=ml26FjTl-C8) — Structures with Prof. H
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/static-determinacy-stability-truss
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1

## Question

Explain static determinacy and stability of the following truss. [Figure: rectangular truss with two panels and cross-bracing: a panel framework with diagonals, supported by a pin at the left bottom and a real hinge at the right bottom.]

## Short answer

For j=6 joints, m=11 members, r=4 reactions, m+r=15 > 2j=12, so it is statically indeterminate to the third degree (external 1, internal2). It is stable because each braced panel is internally rigid and the supports prevent rigid-body translation and rotation.

## Answer

Assumed truss: a rectangular truss with two panels; each panel has a top chord, a bottom chord, vertical members at the ends and a **pair of crossing diagonals** (cross-bracing). So there are 3 joints on top and 3 at the bottom. The left support is a pin (2 reactions) and the right support is a hinge (2 reactions).

#### Count
- Joints $j = 6$
- Members: top chords 2 + bottom chords 2 + verticals 3 + diagonals 4 = $m = 11$
- Reactions $r = 2 + 2 = 4$

#### Static determinacy
For a plane truss, the condition for a determinate truss is $m + r = 2j$.

$$
m + r = 11 + 4 = 15,\qquad 2j = 12
$$

Since $m + r > 2j$, the truss is **statically indeterminate**, with degree of indeterminacy
$$
D_s = (m + r) - 2j = 15 - 12 = 3
$$
Split as external and internal: external degree $= r - 3 = 1$ (the pin plus hinge provide 4 reactions against 3 equilibrium equations), and internal degree $= m - (2j - 3) = 11 - 9 = 2$ (one extra diagonal in each panel).

#### Stability
- Each panel is a braced rectangle (made of triangles by the diagonals), so the truss is internally stable (rigid), with no mechanism.
- The two supports are both pinned, so the reactions are not all parallel or concurrent and the truss cannot move as a rigid body (no translation or rotation).

Hence the truss is **stable but statically indeterminate to the third degree**. (If only one diagonal were used in each panel, $m = 9$, $m + r = 13 > 12$ and the truss would be indeterminate to the first degree, externally only.)

## Check yourself

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

1. Which condition must hold for a plane truss to be statically determinate?
   A. m + r = 2j
   B. m + r = 3j
   C. m = 2j + r
   D. m + j = 2r
   Answer: A. A determinate plane truss satisfies m + r = 2j, where m is members, r reactions and j joints. (Key point: Determinacy criterion)

2. For the two-panel cross-braced truss (j = 6, m = 11, r = 4), what is the degree of static indeterminacy?
   A. 0
   B. 1
   C. 2
   D. 3
   Answer: D. (m + r) - 2j = 15 - 12 = 3, so the truss is indeterminate to the third degree. (Key point: Counting for this truss)

3. How does the answer split the third-degree indeterminacy?
   A. External 3, internal 0
   B. External 2, internal 1
   C. External 0, internal 3
   D. External 1, internal 2
   Answer: D. External degree = r - 3 = 1 (4 reactions against 3 equations); internal degree = m - (2j - 3) = 11 - 9 = 2 (one extra diagonal in each panel). (Key point: External and internal indeterminacy)

4. Why is the truss judged stable although it is indeterminate?
   A. Because m + r > 2j always means stable
   B. Because the braced panels are rigid and the supports prevent rigid-body translation and rotation
   C. Because the right support is a roller
   D. Because the diagonals carry no force
   Answer: B. The diagonals make each panel rigid (no mechanism) and the two pinned supports are neither parallel nor concurrent, so the truss cannot move as a rigid body. (Key point: Stability)

## Related in Strength of Materials

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## Related in other subjects

- [Truss ILDs for BC, CH, HG, bottom chord](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Atruss-ild-bc-ch-hg-bottom) — Theory of Structures I, chapter 5
- [Indeterminacy: frame with hinge and truss](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Adegrees-indeterminacy-frame-and-truss) — Theory of Structures II, chapter 1
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