Stability, Determinacy, and Classification of Structures

Structural Analysis · FE Reference Handbook section

Structural Analysis

4 formulas

10 exam-style examples

~53 min

All Structural Analysis lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Stability, Determinacy, and Classification of Structures within Structural Analysis. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what stability, determinacy, and classification of structures describes physically and when it applies.
State every one of the 4 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: keep kip-ft consistently; EI in kip-in² needs a 1,728 factor.

Lecture

Why this section exists. Stability, Determinacy, and Classification of Structures is the part of Structural Analysis that lets you connect a determinate or one-degree indeterminate structure to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as reactions, an influence-line ordinate, or one deflection. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. keep kip-ft consistently; EI in kip-in² needs a 1,728 factor. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Interior of a steel and glass pedestrian bridge showing the structural framing. — Photo 1. Where this shows up in practice: stability, determinacy, and classification of structures.
Wikimedia Commons, CC BY-SA 4.0

Structural Analysis — Stability, Determinacy, and Classification of Structures: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a determinate or one-degree indeterminate structure. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 4 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

m	Quantity produced by "m = number of members" — read its definition and unit from the handbook line directly above the equation.
r	Quantity produced by "r = number of independent reaction components" — read its definition and unit from the handbook line directly above the equation.
j	Quantity produced by "j = number of joints" — read its definition and unit from the handbook line directly above the equation.
c	Quantity produced by "c = number of condition equations based on known internal moments or forces, such as internal moment of zero" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

at a hinge

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Plane truss determinacy and a chord force by sections — Stability, Determinacy, and Classification of Structures

A symmetric plane truss spans 25 ft in 6 equal panels with a depth of 12 ft and has 11 members and 7 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 31 kip load at midspan.

Given

m = 11 members, j = 7 joints
Span = 25 ft in 6 panels
Depth h = 12 ft
P = 31 kips at midspan

Find

Determinacy check and the bottom-chord force

Start with the thinking

A plane truss is statically determinate when m + 3 = 2j.
One cut through the panel plus moment about the opposite joint gives the chord force directly.

Step-by-step solution

Formula
$d e t er mina cy : m + r = 2 j w i t h r = 3$
Substituting
$11 + 3 = 14 v er s u s 2 j = 14 \to d e t er mina t e$
Reaction
$R = P /2 = 31/2 = 15.5 k i p s$
Panel length
$s = 25/6 = 4.17 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 15.5 (4.17) /12 = 5.38 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 5.38 kips tension

Why the other options are there

64.6 kips (never divided by the depth)
31.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Example 2

Plane truss determinacy and a chord force by sections — Stability, Determinacy, and Classification of Structures (2)

A symmetric plane truss spans 56 ft in 8 equal panels with a depth of 9 ft and has 21 members and 12 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 34 kip load at midspan.

Given

m = 21 members, j = 12 joints
Span = 56 ft in 8 panels
Depth h = 9 ft
P = 34 kips at midspan

Find

Determinacy check and the bottom-chord force

Start with the thinking

A plane truss is statically determinate when m + 3 = 2j.
One cut through the panel plus moment about the opposite joint gives the chord force directly.

Step-by-step solution

Formula
$d e t er mina cy : m + r = 2 j w i t h r = 3$
Substituting
$21 + 3 = 24 v er s u s 2 j = 24 \to d e t er mina t e$
Reaction
$R = P /2 = 34/2 = 17.0 k i p s$
Panel length
$s = 56/8 = 7.00 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 17.0 (7.00) /9 = 13.22 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 13.22 kips tension

Why the other options are there

119.0 kips (never divided by the depth)
34.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Example 3

Static determinacy of a planar truss — Stability, Determinacy, and Classification of Structures (3)

A planar truss has 22 members, 12 joints and 4 reaction components. Classify it.

Given

m = 22
j = 12
r = 4

Find

Degree of static indeterminacy

Start with the thinking

Compare available equations (2j) with unknowns (m + r).
A negative result means a mechanism, not redundancy.

Step-by-step solution

Criterion
$d e g r ee = m + r - 2 j$
Substituting
$d e g r ee = 22 + 4 - 2 (12) = 2$
Classification — indeterminate to degree 2

Answer: Indeterminate to degree 2

Why the other options are there

Degree -2 (reactions omitted)
Degree 14 (only one equation per joint used)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Example 4

Plane truss determinacy and a chord force by sections — Stability, Determinacy, and Classification of Structures (3)

A symmetric plane truss spans 26 ft in 8 equal panels with a depth of 13 ft and has 21 members and 12 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 38 kip load at midspan.

Given

m = 21 members, j = 12 joints
Span = 26 ft in 8 panels
Depth h = 13 ft
P = 38 kips at midspan

Find

Determinacy check and the bottom-chord force

Start with the thinking

A plane truss is statically determinate when m + 3 = 2j.
One cut through the panel plus moment about the opposite joint gives the chord force directly.

Step-by-step solution

Formula
$d e t er mina cy : m + r = 2 j w i t h r = 3$
Substituting
$21 + 3 = 24 v er s u s 2 j = 24 \to d e t er mina t e$
Reaction
$R = P /2 = 38/2 = 19.0 k i p s$
Panel length
$s = 26/8 = 3.25 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 19.0 (3.25) /13 = 4.75 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 4.75 kips tension

Why the other options are there

61.8 kips (never divided by the depth)
38.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Example 5

Static determinacy of a planar truss — Stability, Determinacy, and Classification of Structures (4)

A planar truss has 24 members, 13 joints and 3 reaction components. Classify it.

Given

m = 24
j = 13
r = 3

Find

Degree of static indeterminacy

Start with the thinking

Compare available equations (2j) with unknowns (m + r).
A negative result means a mechanism, not redundancy.

Step-by-step solution

Criterion
$d e g r ee = m + r - 2 j$
Substituting
$d e g r ee = 24 + 3 - 2 (13) = 1$
Classification — indeterminate to degree 1

Answer: Indeterminate to degree 1

Why the other options are there

Degree -2 (reactions omitted)
Degree 14 (only one equation per joint used)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Example 6

Plane truss determinacy and a chord force by sections — Stability, Determinacy, and Classification of Structures (4)

A symmetric plane truss spans 29 ft in 4 equal panels with a depth of 15 ft and has 11 members and 7 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 12 kip load at midspan.

Given

m = 11 members, j = 7 joints
Span = 29 ft in 4 panels
Depth h = 15 ft
P = 12 kips at midspan

Find

Determinacy check and the bottom-chord force

Start with the thinking

A plane truss is statically determinate when m + 3 = 2j.
One cut through the panel plus moment about the opposite joint gives the chord force directly.

Step-by-step solution

Formula
$d e t er mina cy : m + r = 2 j w i t h r = 3$
Substituting
$11 + 3 = 14 v er s u s 2 j = 14 \to d e t er mina t e$
Reaction
$R = P /2 = 12/2 = 6.0 k i p s$
Panel length
$s = 29/4 = 7.25 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 6.0 (7.25) /15 = 2.90 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 2.90 kips tension

Why the other options are there

43.5 kips (never divided by the depth)
12.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Example 7

Static determinacy of a planar truss — Stability, Determinacy, and Classification of Structures (5)

A planar truss has 18 members, 10 joints and 3 reaction components. Classify it.

Given

m = 18
j = 10
r = 3

Find

Degree of static indeterminacy

Start with the thinking

Compare available equations (2j) with unknowns (m + r).
A negative result means a mechanism, not redundancy.

Step-by-step solution

Criterion
$d e g r ee = m + r - 2 j$
Substituting
$d e g r ee = 18 + 3 - 2 (10) = 1$
Classification — indeterminate to degree 1

Answer: Indeterminate to degree 1

Why the other options are there

Degree -2 (reactions omitted)
Degree 11 (only one equation per joint used)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Example 8

Plane truss determinacy and a chord force by sections — Stability, Determinacy, and Classification of Structures (5)

A symmetric plane truss spans 32 ft in 8 equal panels with a depth of 16 ft and has 17 members and 10 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 10 kip load at midspan.

Given

m = 17 members, j = 10 joints
Span = 32 ft in 8 panels
Depth h = 16 ft
P = 10 kips at midspan

Find

Determinacy check and the bottom-chord force

Start with the thinking

A plane truss is statically determinate when m + 3 = 2j.
One cut through the panel plus moment about the opposite joint gives the chord force directly.

Step-by-step solution

Formula
$d e t er mina cy : m + r = 2 j w i t h r = 3$
Substituting
$17 + 3 = 20 v er s u s 2 j = 20 \to d e t er mina t e$
Reaction
$R = P /2 = 10/2 = 5.0 k i p s$
Panel length
$s = 32/8 = 4.00 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 5.0 (4.00) /16 = 1.25 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 1.25 kips tension

Why the other options are there

20.0 kips (never divided by the depth)
10.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Example 9

Static determinacy of a planar truss — Stability, Determinacy, and Classification of Structures (6)

A planar truss has 19 members, 11 joints and 4 reaction components. Classify it.

Given

m = 19
j = 11
r = 4

Find

Degree of static indeterminacy

Start with the thinking

Compare available equations (2j) with unknowns (m + r).
A negative result means a mechanism, not redundancy.

Step-by-step solution

Criterion
$d e g r ee = m + r - 2 j$
Substituting
$d e g r ee = 19 + 4 - 2 (11) = 1$
Classification — indeterminate to degree 1

Answer: Indeterminate to degree 1

Why the other options are there

Degree -3 (reactions omitted)
Degree 12 (only one equation per joint used)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Example 10

Plane truss determinacy and a chord force by sections — Stability, Determinacy, and Classification of Structures (6)

A symmetric plane truss spans 27 ft in 6 equal panels with a depth of 11 ft and has 15 members and 9 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 24 kip load at midspan.

Given

m = 15 members, j = 9 joints
Span = 27 ft in 6 panels
Depth h = 11 ft
P = 24 kips at midspan

Find

Determinacy check and the bottom-chord force

Start with the thinking

A plane truss is statically determinate when m + 3 = 2j.
One cut through the panel plus moment about the opposite joint gives the chord force directly.

Step-by-step solution

Formula
$d e t er mina cy : m + r = 2 j w i t h r = 3$
Substituting
$15 + 3 = 18 v er s u s 2 j = 18 \to d e t er mina t e$
Reaction
$R = P /2 = 24/2 = 12.0 k i p s$
Panel length
$s = 27/6 = 4.50 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 12.0 (4.50) /11 = 4.91 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 4.91 kips tension

Why the other options are there

54.0 kips (never divided by the depth)
24.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Stability, Determinacy, and Classification of Structures

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a determinate or one-degree indeterminate structure, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Stability, Determinacy, and Classification of Structures contains 4 relations; you must be able to find this page in under 15 seconds.
Exam style: reactions, an influence-line ordinate, or one deflection.
Unit rule: keep kip-ft consistently; EI in kip-in² needs a 1,728 factor.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

keep kip-ft consistently; EI in kip-in² needs a 1,728 factor
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.