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How do engineers find the internal force in each member of a loaded truss?

Analyse a statically determinate pin-jointed truss using the method of joints to find the magnitude and nature (tension or compression) of the force in each member

A QCE Engineering Unit 3 answer on analysing pin-jointed trusses. Covers two-force members, the assumption of pin joints, the method of joints, and how to classify each member force as tension or compression with worked arithmetic.

Generated by Claude Opus 4.86 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to take a pin-jointed truss carrying known loads and find the internal force in every member, stating whether each is in tension or compression. The standard tool is the method of joints: isolate each pin, treat it as a particle in equilibrium, and solve. This is the structural analysis step that turns a bridge or roof frame into a set of member forces you can later check against material limits.

The answer

What makes a truss special

A truss is an assembly of straight members connected at their ends by frictionless pins, with all external loads applied at the joints. These assumptions matter because they make every member a two-force member: a body with forces at only two points must carry a force directed along the line joining them. So each member is in pure tension (pulling its end joints together) or pure compression (pushing them apart), with no bending. That is what makes hand analysis possible.

A truss is statically determinate when the equilibrium equations alone are enough to find all member forces. The check is:

m + r = 2j

where $m$ is the number of members, $r$ the number of reaction components, and $j$ the number of joints. If this holds and the truss is stable, the method of joints will solve it.

The method of joints

The method treats each pin as a particle in equilibrium. A particle has no moment equation, so each joint gives exactly two equations:

\sum F_x = 0 \qquad \sum F_y = 0

The procedure:

Find the support reactions from whole-structure equilibrium ( $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum M = 0$ ).
Pick a joint with no more than two unknown member forces.
Draw the joint's free-body diagram. Assume each unknown member pulls away from the joint (tension positive).
Resolve member forces into components using the member angles and solve.
Carry the now-known forces to adjacent joints and repeat.

A positive result confirms the member is in tension. A negative result means the true force points the other way, so the member is in compression. Keep the sign convention consistent throughout.

Reading tension and compression

Tension members pull their joints inward and tend to stretch; compression members push their joints outward and tend to shorten. Compression members also risk buckling, so engineers often size them more generously than tension members of the same force. Labelling each solved member as T or C is part of a complete answer.

Force in a loaded diagonal

Find the member forces at a loaded joint

A joint C in a truss has three members meeting it: a horizontal member CB to the left, a horizontal member CD to the right, and a diagonal member CA rising to the left at $30^\circ$ above the horizontal. A vertical downward load of $5.0\,\text{kN}$ acts at C. Member CD carries no vertical component. Find the forces in CA and CD, assuming CB is already known to be zero for this joint check.

Take tension as positive (arrows pull away from C). The diagonal CA points up-left, so its components at C are $F_{CA}\cos 30^\circ$ (horizontal) and $F_{CA}\sin 30^\circ$ (vertical).

Vertical equilibrium ( $\sum F_y = 0$ ):

F_{CA}\sin 30^\circ - 5.0 = 0

F_{CA} = \frac{5.0}{0.5} = 10.0\,\text{kN}

The positive sign means CA is in tension at $10.0\,\text{kN}$ .

Horizontal equilibrium ( $\sum F_x = 0$ ), taking right as positive:

F_{CD} - F_{CA}\cos 30^\circ = 0

F_{CD} = 10.0 \times 0.866 = 8.66\,\text{kN}

The positive sign means CD is in tension at $8.66\,\text{kN}$ . Check: $10.0 \times \sin 30^\circ = 5.0\,\text{kN}$ exactly balances the load, confirming the arithmetic.

Why this matters for civil structures

Truss analysis is how roof frames, bridge spans and transmission towers are designed. Each member force feeds straight into a stress calculation ( $\sigma = F/A$ ) to choose a cross-section, and compression members get an extra buckling check. A single mis-signed force can lead to an undersized member, so the tension or compression label is as important as the magnitude.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20227 marksA pin-jointed joint connects a horizontal member and a diagonal member that rises at

30^\circ

above the horizontal. An external load of

5.0\ \text{kN}

pulls vertically downward at the joint. Using the method of joints, determine the force in the diagonal member and state whether it is in tension or compression.

Show worked answer →

A 7 mark determine question rewards resolving at the joint and classifying the member.

Resolve vertically at the joint. Only the diagonal has a vertical component, which must balance the $5.0\ \text{kN}$ downward load: $F_{\text{diag}} \sin 30^\circ = 5.0$ , so $F_{\text{diag}} = \dfrac{5.0}{\sin 30^\circ} = \dfrac{5.0}{0.5} = 10\ \text{kN}$ .

The diagonal must pull upward on the joint to hold up the load, which means the joint pulls down on the member, so the member is in tension.

Markers reward the vertical equilibrium equation, the magnitude of $10\ \text{kN}$ , and the correct tension or compression sense justified by the direction the member acts on the joint.

QCAA 20234 marksExplain why a truss is analysed by assuming frictionless pin joints and loads applied only at the joints, and what this assumption means for the members.

Show worked answer →

A 4 mark explain answer needs the consequence of the idealisation.

Assuming pin joints means the joints cannot carry moments, so each member carries force only along its own axis. With loads applied only at the joints, every member becomes a two-force member, in pure tension or pure compression with no bending. This is what allows the method of joints (resolving forces at each pin) to determine every member force.

Markers reward stating that pin joints plus joint loading make each member a two-force (axial only) member, enabling the equilibrium analysis.