Plane Truss

Structural Analysis · FE Reference Handbook section

Structural Analysis

1 formulas

10 exam-style examples

~47 min

All Structural Analysis lectures

Learning objectives

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

This chapter section covers Plane Truss 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 plane truss describes physically and when it applies.
State every one of the 1 relation 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. Plane Truss 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: plane truss.
Wikimedia Commons, CC BY-SA 4.0

Structural Analysis — Plane Truss: 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 1 relation 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 + r	Quantity produced by "m + r = 2j Stable and statically determinate" — 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.

Static Analysis Classification
m + r < 2j Unstable
m + r > 2j Stable and statically indeterminate

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

Truss member force by the method of joints

A symmetric pin-jointed truss of span 12 m and height 4 m carries 60 kN at the apex. Find the force in each of the two inclined top members.

Given

Span = 12 m, height = 4 m
Apex load = 60 kN
Symmetric geometry

Find

Force in the inclined members

Start with the thinking

Symmetry means each inclined member carries half the vertical load.
Resolve at the apex joint using the member slope.

Figure for Truss member force by the method of joints

Step-by-step solution

Member length
$L = (6^{2} + 4^{2}) = 7.21 m$
Vertical component per member
$F v = 60/2 = 30.0 k N$
Slope ratio
$L / h e i g h t = 7.21/4 = 1.803$
Member force
$F = F v (L / h e i g h t) = 30.0 (1.803)$
Result
$F = 54.1 k N co m p r ess i o nin e a c hin c l in e d m e mb er$

Answer: 54.1 kN compression

Why the other options are there

30.0 kN (vertical component reported as member force)
45.0 kN (horizontal projection used)

Reference: FE Reference Handbook — Statics — Trusses

Example 2

Plane truss determinacy and a chord force by sections — Plane Truss

A symmetric plane truss spans 39 ft in 4 equal panels with a depth of 8 ft and has 9 members and 6 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 19 kip load at midspan.

Given

m = 9 members, j = 6 joints
Span = 39 ft in 4 panels
Depth h = 8 ft
P = 19 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
$9 + 3 = 12 v er s u s 2 j = 12 \to d e t er mina t e$
Reaction
$R = P /2 = 19/2 = 9.5 k i p s$
Panel length
$s = 39/4 = 9.75 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 9.5 (9.75) /8 = 11.58 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 11.58 kips tension

Why the other options are there

92.6 kips (never divided by the depth)
19.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Plane Truss

Example 3

Plane truss determinacy and a chord force by sections — Plane Truss (2)

A symmetric plane truss spans 27 ft in 4 equal panels with a depth of 14 ft and has 9 members and 6 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 21 kip load at midspan.

Given

m = 9 members, j = 6 joints
Span = 27 ft in 4 panels
Depth h = 14 ft
P = 21 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
$9 + 3 = 12 v er s u s 2 j = 12 \to d e t er mina t e$
Reaction
$R = P /2 = 21/2 = 10.5 k i p s$
Panel length
$s = 27/4 = 6.75 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 10.5 (6.75) /14 = 5.06 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 5.06 kips tension

Why the other options are there

70.9 kips (never divided by the depth)
21.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Plane Truss

Example 4

Plane truss determinacy and a chord force by sections — Plane Truss (3)

A symmetric plane truss spans 58 ft in 8 equal panels with a depth of 10 ft and has 11 members and 7 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 33 kip load at midspan.

Given

m = 11 members, j = 7 joints
Span = 58 ft in 8 panels
Depth h = 10 ft
P = 33 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
$R = P /2 = 33/2 = 16.5 k i p s$
Panel length
$s = 58/8 = 7.25 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 16.5 (7.25) /10 = 11.96 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 11.96 kips tension

Why the other options are there

119.6 kips (never divided by the depth)
33.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Plane Truss

Example 5

Plane truss determinacy and a chord force by sections — Plane Truss (4)

A symmetric plane truss spans 47 ft in 6 equal panels with a depth of 15 ft and has 21 members and 12 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 37 kip load at midspan.

Given

m = 21 members, j = 12 joints
Span = 47 ft in 6 panels
Depth h = 15 ft
P = 37 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
$R = P /2 = 37/2 = 18.5 k i p s$
Panel length
$s = 47/6 = 7.83 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 18.5 (7.83) /15 = 9.66 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 9.66 kips tension

Why the other options are there

144.9 kips (never divided by the depth)
37.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Plane Truss

Example 6

Plane truss determinacy and a chord force by sections — Plane Truss (5)

A symmetric plane truss spans 29 ft in 8 equal panels with a depth of 12 ft and has 17 members and 10 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 31 kip load at midspan.

Given

m = 17 members, j = 10 joints
Span = 29 ft in 8 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
$17 + 3 = 20 v er s u s 2 j = 20 \to d e t er mina t e$
Reaction
$R = P /2 = 31/2 = 15.5 k i p s$
Panel length
$s = 29/8 = 3.63 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 15.5 (3.63) /12 = 4.68 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 4.68 kips tension

Why the other options are there

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

Reference: FE Reference Handbook — Structural Analysis → Plane Truss

Example 7

Plane truss determinacy and a chord force by sections — Plane Truss (6)

A symmetric plane truss spans 27 ft in 4 equal panels with a depth of 15 ft and has 9 members and 6 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 15 kip load at midspan.

Given

m = 9 members, j = 6 joints
Span = 27 ft in 4 panels
Depth h = 15 ft
P = 15 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
$9 + 3 = 12 v er s u s 2 j = 12 \to d e t er mina t e$
Reaction
$R = P /2 = 15/2 = 7.5 k i p s$
Panel length
$s = 27/4 = 6.75 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 7.5 (6.75) /15 = 3.38 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 3.38 kips tension

Why the other options are there

50.6 kips (never divided by the depth)
15.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Plane Truss

Example 8

Plane truss determinacy and a chord force by sections — Plane Truss (7)

A symmetric plane truss spans 45 ft in 4 equal panels with a depth of 15 ft and has 9 members and 6 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 15 kip load at midspan.

Given

m = 9 members, j = 6 joints
Span = 45 ft in 4 panels
Depth h = 15 ft
P = 15 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
$9 + 3 = 12 v er s u s 2 j = 12 \to d e t er mina t e$
Reaction
$R = P /2 = 15/2 = 7.5 k i p s$
Panel length
$s = 45/4 = 11.25 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 7.5 (11.25) /15 = 5.63 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 5.63 kips tension

Why the other options are there

84.4 kips (never divided by the depth)
15.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Plane Truss

Example 9

Plane truss determinacy and a chord force by sections — Plane Truss (8)

A symmetric plane truss spans 53 ft in 8 equal panels with a depth of 11 ft and has 21 members and 12 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 25 kip load at midspan.

Given

m = 21 members, j = 12 joints
Span = 53 ft in 8 panels
Depth h = 11 ft
P = 25 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
$R = P /2 = 25/2 = 12.5 k i p s$
Panel length
$s = 53/8 = 6.63 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 12.5 (6.63) /11 = 7.53 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 7.53 kips tension

Why the other options are there

82.8 kips (never divided by the depth)
25.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Plane Truss

Example 10

Plane truss determinacy and a chord force by sections — Plane Truss (9)

A symmetric plane truss spans 52 ft in 8 equal panels with a depth of 8 ft and has 17 members and 10 joints. Check static determinacy, then find the bottom-chord force in the first panel for a single 29 kip load at midspan.

Given

m = 17 members, j = 10 joints
Span = 52 ft in 8 panels
Depth h = 8 ft
P = 29 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
$R = P /2 = 29/2 = 14.5 k i p s$
Panel length
$s = 52/8 = 6.50 f t$
Formula — ΣM at the top joint: F_chord × h = R × s
Substituting
$F = 14.5 (6.50) /8 = 11.78 k i p s (t e n s i o n)$

Answer: Truss is determinate; bottom chord F = 11.78 kips tension

Why the other options are there

94.3 kips (never divided by the depth)
29.0 kips (used the whole load)

Reference: FE Reference Handbook — Structural Analysis → Plane Truss

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

Plane Truss contains 1 relation; 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.

Plane Truss

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

m+r=2jStableandstaticallydeterminate

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section