Plane Truss: Method of Joints

Statics · FE Reference Handbook section

Statics

3 formulas

10 exam-style examples

~51 min

All Statics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The method consists of solving for the forces in the members by writing the two equilibrium equations for each joint

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

At a truss joint, a 11 kip downward load is carried by a diagonal inclined 57° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 11 kip$
$\theta = 57^{\circ}$

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
$F_d = 11/\sin 57^{\circ} = 13.12 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 13.12·cos 57° = 7.14 kip (compression)

Answer:

F_diagonal = 13.12 kip, F_chord = 7.14 kip

Why the other options are there

9.23 kip (sine multiplied instead of divided)
11 kip (load taken directly as the member force)

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

Example 2

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

At a truss joint, a 23 kip downward load is carried by a diagonal inclined 59° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 23 kip$
$\theta = 59^{\circ}$

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
$F_d = 23/\sin 59^{\circ} = 26.83 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 26.83·cos 59° = 13.82 kip (compression)

Answer:

F_diagonal = 26.83 kip, F_chord = 13.82 kip

Why the other options are there

19.71 kip (sine multiplied instead of divided)
23 kip (load taken directly as the member force)

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

Example 3

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

At a truss joint, a 29 kip downward load is carried by a diagonal inclined 59° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 29 kip$
$\theta = 59^{\circ}$

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
$F_d = 29/\sin 59^{\circ} = 33.83 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 33.83·cos 59° = 17.42 kip (compression)

Answer:

F_diagonal = 33.83 kip, F_chord = 17.42 kip

Why the other options are there

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

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

Example 4

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

At a truss joint, a 22 kip downward load is carried by a diagonal inclined 35° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 22 kip$
$\theta = 35^{\circ}$

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
$F_d = 22/\sin 35^{\circ} = 38.36 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 38.36·cos 35° = 31.42 kip (compression)

Answer:

F_diagonal = 38.36 kip, F_chord = 31.42 kip

Why the other options are there

12.62 kip (sine multiplied instead of divided)
22 kip (load taken directly as the member force)

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

Example 5

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

At a truss joint, a 40 kip downward load is carried by a diagonal inclined 30° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 40 kip$
$\theta = 30^{\circ}$

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
$F_d = 40/\sin 30^{\circ} = 80.00 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 80.00·cos 30° = 69.28 kip (compression)

Answer:

F_diagonal = 80.00 kip, F_chord = 69.28 kip

Why the other options are there

20.00 kip (sine multiplied instead of divided)
40 kip (load taken directly as the member force)

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

Example 6

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

At a truss joint, a 18 kip downward load is carried by a diagonal inclined 36° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 18 kip$
$\theta = 36^{\circ}$

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
$F_d = 18/\sin 36^{\circ} = 30.62 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 30.62·cos 36° = 24.77 kip (compression)

Answer:

F_diagonal = 30.62 kip, F_chord = 24.77 kip

Why the other options are there

10.58 kip (sine multiplied instead of divided)
18 kip (load taken directly as the member force)

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

Example 7

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

At a truss joint, a 32 kip downward load is carried by a diagonal inclined 57° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 32 kip$
$\theta = 57^{\circ}$

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
$F_d = 32/\sin 57^{\circ} = 38.16 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 38.16·cos 57° = 20.78 kip (compression)

Answer:

F_diagonal = 38.16 kip, F_chord = 20.78 kip

Why the other options are there

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

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

Example 8

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

At a truss joint, a 33 kip downward load is carried by a diagonal inclined 44° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 33 kip$
$\theta = 44^{\circ}$

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
$F_d = 33/\sin 44^{\circ} = 47.51 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 47.51·cos 44° = 34.17 kip (compression)

Answer:

F_diagonal = 47.51 kip, F_chord = 34.17 kip

Why the other options are there

22.92 kip (sine multiplied instead of divided)
33 kip (load taken directly as the member force)

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

Example 9

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

At a truss joint, a 25 kip downward load is carried by a diagonal inclined 35° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 25 kip$
$\theta = 35^{\circ}$

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
$F_d = 25/\sin 35^{\circ} = 43.59 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 43.59·cos 35° = 35.70 kip (compression)

Answer:

F_diagonal = 43.59 kip, F_chord = 35.70 kip

Why the other options are there

14.34 kip (sine multiplied instead of divided)
25 kip (load taken directly as the member force)

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

Example 10

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

At a truss joint, a 29 kip downward load is carried by a diagonal inclined 36° from horizontal and a horizontal chord. Find both member forces.

Given

$P = 29 kip$
$\theta = 36^{\circ}$

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
$F_d = 29/\sin 36^{\circ} = 49.34 kip (tension)$
Horizontal equilibrium — ΣF_x = 0: F_c = F_d·cos θ
Substituting — F_c = 49.34·cos 36° = 39.92 kip (compression)

Answer:

F_diagonal = 49.34 kip, F_chord = 39.92 kip

Why the other options are there

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

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