Plane Truss: Method of Sections

Statics · FE Reference Handbook section

Statics

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10 exam-style examples

~45 min

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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

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

$P = 30 kip$
$\theta = 49^{\circ}$

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
$F_d = 30/\sin 49^{\circ} = 39.75 kip (tension)$
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

$P = 32 kip$
$\theta = 47^{\circ}$

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
$F_d = 32/\sin 47^{\circ} = 43.75 kip (tension)$
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

$P = 12 kip$
$\theta = 50^{\circ}$

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
$F_d = 12/\sin 50^{\circ} = 15.66 kip (tension)$
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

$P = 34 kip$
$\theta = 37^{\circ}$

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
$F_d = 34/\sin 37^{\circ} = 56.50 kip (tension)$
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

$P = 20 kip$
$\theta = 37^{\circ}$

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
$F_d = 20/\sin 37^{\circ} = 33.23 kip (tension)$
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

$P = 15 kip$
$\theta = 50^{\circ}$

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
$F_d = 15/\sin 50^{\circ} = 19.58 kip (tension)$
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

$P = 32 kip$
$\theta = 58^{\circ}$

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
$F_d = 32/\sin 58^{\circ} = 37.73 kip (tension)$
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

$P = 20 kip$
$\theta = 45^{\circ}$

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
$F_d = 20/\sin 45^{\circ} = 28.28 kip (tension)$
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

$P = 35 kip$
$\theta = 48^{\circ}$

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
$F_d = 35/\sin 48^{\circ} = 47.10 kip (tension)$
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

$P = 29 kip$
$\theta = 41^{\circ}$

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
$F_d = 29/\sin 41^{\circ} = 44.20 kip (tension)$
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