Plane Truss: Method of Sections

Statics · FE Reference Handbook section

Statics

0 formulas

10 exam-style examples

~45 min

All Statics lectures

Learning objectives

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

This chapter section covers Plane Truss: Method of Sections within Statics. 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: method of sections describes physically and when it applies.
State every one of the 0 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 force in lbf or kN and distance in ft or m consistently.

Lecture

Why this section exists. Plane Truss: Method of Sections is the part of Statics that lets you connect a determinate frame, truss or beam in equilibrium 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 one free body, three equilibrium equations, one unknown reported. 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 force in lbf or kN and distance in ft or m consistently. 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: method of sections.
Wikimedia Commons, CC BY-SA 4.0

Statics — Plane Truss: Method of Sections: 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 frame, truss or beam in equilibrium. 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 0 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The method consists of drawing a free-body diagram of a portion of the truss in such a way that the unknown truss member
force is exposed as an external force.

Core formulas for this FE topic

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

This section is conceptual; there are no equations to memorise.

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 forces at a loaded joint — Plane Truss: Method of Sections

At a truss joint, a 30 kip downward load is carried by a diagonal inclined 49° from horizontal and a horizontal chord. Find both member forces.

Given

P = 30 kip
θ = 49°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 30/ sin 4 9^{\circ} = 39.75 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 39.75·cos 49° = 26.08 kip (compression)

Answer: F_diagonal = 39.75 kip, F_chord = 26.08 kip

Why the other options are there

22.64 kip (sine multiplied instead of divided)
30 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

Example 2

Truss member forces at a loaded joint — Plane Truss: Method of Sections (2)

At a truss joint, a 32 kip downward load is carried by a diagonal inclined 47° from horizontal and a horizontal chord. Find both member forces.

Given

P = 32 kip
θ = 47°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 32/ sin 4 7^{\circ} = 43.75 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 43.75·cos 47° = 29.84 kip (compression)

Answer: F_diagonal = 43.75 kip, F_chord = 29.84 kip

Why the other options are there

23.40 kip (sine multiplied instead of divided)
32 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

Example 3

Truss member forces at a loaded joint — Plane Truss: Method of Sections (3)

At a truss joint, a 12 kip downward load is carried by a diagonal inclined 50° from horizontal and a horizontal chord. Find both member forces.

Given

P = 12 kip
θ = 50°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 12/ sin 5 0^{\circ} = 15.66 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 15.66·cos 50° = 10.07 kip (compression)

Answer: F_diagonal = 15.66 kip, F_chord = 10.07 kip

Why the other options are there

9.19 kip (sine multiplied instead of divided)
12 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

Example 4

Truss member forces at a loaded joint — Plane Truss: Method of Sections (4)

At a truss joint, a 34 kip downward load is carried by a diagonal inclined 37° from horizontal and a horizontal chord. Find both member forces.

Given

P = 34 kip
θ = 37°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 34/ sin 3 7^{\circ} = 56.50 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 56.50·cos 37° = 45.12 kip (compression)

Answer: F_diagonal = 56.50 kip, F_chord = 45.12 kip

Why the other options are there

20.46 kip (sine multiplied instead of divided)
34 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

Example 5

Truss member forces at a loaded joint — Plane Truss: Method of Sections (5)

At a truss joint, a 20 kip downward load is carried by a diagonal inclined 37° from horizontal and a horizontal chord. Find both member forces.

Given

P = 20 kip
θ = 37°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 20/ sin 3 7^{\circ} = 33.23 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 33.23·cos 37° = 26.54 kip (compression)

Answer: F_diagonal = 33.23 kip, F_chord = 26.54 kip

Why the other options are there

12.04 kip (sine multiplied instead of divided)
20 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

Example 6

Truss member forces at a loaded joint — Plane Truss: Method of Sections (6)

At a truss joint, a 15 kip downward load is carried by a diagonal inclined 50° from horizontal and a horizontal chord. Find both member forces.

Given

P = 15 kip
θ = 50°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 15/ sin 5 0^{\circ} = 19.58 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 19.58·cos 50° = 12.59 kip (compression)

Answer: F_diagonal = 19.58 kip, F_chord = 12.59 kip

Why the other options are there

11.49 kip (sine multiplied instead of divided)
15 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

Example 7

Truss member forces at a loaded joint — Plane Truss: Method of Sections (7)

At a truss joint, a 32 kip downward load is carried by a diagonal inclined 58° from horizontal and a horizontal chord. Find both member forces.

Given

P = 32 kip
θ = 58°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 32/ sin 5 8^{\circ} = 37.73 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 37.73·cos 58° = 20.00 kip (compression)

Answer: F_diagonal = 37.73 kip, F_chord = 20.00 kip

Why the other options are there

27.14 kip (sine multiplied instead of divided)
32 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

Example 8

Truss member forces at a loaded joint — Plane Truss: Method of Sections (8)

At a truss joint, a 20 kip downward load is carried by a diagonal inclined 45° from horizontal and a horizontal chord. Find both member forces.

Given

P = 20 kip
θ = 45°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 20/ sin 4 5^{\circ} = 28.28 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 28.28·cos 45° = 20.00 kip (compression)

Answer: F_diagonal = 28.28 kip, F_chord = 20.00 kip

Why the other options are there

14.14 kip (sine multiplied instead of divided)
20 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

Example 9

Truss member forces at a loaded joint — Plane Truss: Method of Sections (9)

At a truss joint, a 35 kip downward load is carried by a diagonal inclined 48° from horizontal and a horizontal chord. Find both member forces.

Given

P = 35 kip
θ = 48°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 35/ sin 4 8^{\circ} = 47.10 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 47.10·cos 48° = 31.51 kip (compression)

Answer: F_diagonal = 47.10 kip, F_chord = 31.51 kip

Why the other options are there

26.01 kip (sine multiplied instead of divided)
35 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

Example 10

Truss member forces at a loaded joint — Plane Truss: Method of Sections (10)

At a truss joint, a 29 kip downward load is carried by a diagonal inclined 41° from horizontal and a horizontal chord. Find both member forces.

Given

P = 29 kip
θ = 41°

Find

Diagonal and chord forces

Start with the thinking

Two unknowns at a joint means two equilibrium equations are enough.
Resolve the diagonal into components before summing.

Step-by-step solution

Vertical equilibrium — ΣF_y = 0: F_d·sin θ = P
Substituting
$F_{d} = 29/ sin 4 1^{\circ} = 44.20 k i p (t e n s i o n)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 44.20·cos 41° = 33.36 kip (compression)

Answer: F_diagonal = 44.20 kip, F_chord = 33.36 kip

Why the other options are there

19.03 kip (sine multiplied instead of divided)
29 kip (load taken directly as the member force)

Reference: FE Reference Handbook — Statics → Plane Truss: Method of Sections

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 frame, truss or beam in equilibrium, 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: Method of Sections contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: one free body, three equilibrium equations, one unknown reported.
Unit rule: keep force in lbf or kN and distance in ft or m consistently.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

keep force in lbf or kN and distance in ft or m consistently
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.