What are the assumptions in truss analysis?

Members are connected by frictionless pins, loads act only at joints, members carry only axial force (tension or compression), and member weights are neglected or applied at joints. These assumptions make every member a two-force member.

When should I use method of joints versus method of sections?

Use the method of joints to find forces in all or most members, working joint by joint. Use the method of sections when you need the force in only a few specific members, by cutting through them and applying equilibrium to one part of the truss.

What is a zero-force member?

A zero-force member carries no load under a given loading. Two rules identify them: at an unloaded two-member joint both members are zero-force, and at an unloaded three-member joint where two members are collinear, the third is zero-force. They provide stability and handle other load cases.

Analysis of Trusses — Method of Joints and Method of Sections with Examples

Trusses — roofs, bridges, towers, cranes — carry load through triangulated members that act in pure tension or compression. Because they are statically determinate, hand analysis is exact and fast. This article covers both classic methods with full worked examples.

Assumptions of Truss Analysis

Joints are frictionless pins (no moment transfer).
Loads and reactions act only at joints.
Members are straight and carry only axial force — tension (pulling the joint) or compression (pushing the joint).
Self-weight is neglected or lumped at joints.

A truss is statically determinate if m + r = 2j (m = members, r = reactions, j = joints). If m + r > 2j it is indeterminate; if less, it is a mechanism (unstable).

Method of Joints

Each pin joint is in equilibrium under concurrent forces, giving two equations (ΣF_x = 0, ΣF_y = 0) per joint. Start at a joint with no more than two unknowns, solve, and march through the truss.

Worked Example — Method of Joints (Cantilever Bracket Truss)

A bracket truss is fixed to a wall at joints A (top) and B (bottom), 3 m apart vertically. Joint C projects 4 m horizontally out from B at the same level as B. Member BC is horizontal (4 m), member AC is the diagonal (length √(3²+4²) = 5 m). A vertical load of 12 kN hangs at C. Find the member forces.

At joint C the diagonal CA rises 3 vertical to 4 horizontal, so sinθ = 3/5 = 0.6, cosθ = 4/5 = 0.8.

ΣF_y = 0: vertical component of CA must carry 12 kN → F_CA × 0.6 = 12 → F_CA = 20 kN (tension)
ΣF_x = 0: horizontal pull of CA (left) = 20 × 0.8 = 16 kN, balanced by CB → F_CB = 16 kN (compression)

The diagonal is the tie (tension); the horizontal member is the strut (compression) — exactly what intuition expects for a wall bracket.

Method of Sections

To find the force in just a few members, cut an imaginary section through them (no more than three unknown members) and apply the three equilibrium equations to either side. Taking moments about the intersection of two unknowns isolates the third in a single equation.

Worked Example — Method of Sections

A simply supported parallel-chord truss spans 8 m with four 2 m panels, carrying 10 kN at each interior bottom-chord joint (at 2, 4, 6 m). Panel height = 2 m. Find the force in the top chord member over the 2–4 m panel.

Total load = 30 kN; by symmetry each support reaction = 15 kN.
Cut vertically through the panel between x = 2 m and x = 4 m, keep the left part.
Take moments about the bottom-chord joint at x = 4 m (where the diagonal and bottom chord meet), so only the top chord force has a lever arm of 2 m (the height):
ΣM = 0: 15(4) − 10(2) − F_top(2) = 0 → 60 − 20 = 2 F_top → F_top = 20 kN (compression)

One equation, one member — the power of the sections method.

Zero-Force Members

At an unloaded joint with only two non-collinear members, both are zero-force.
At an unloaded joint with three members where two are collinear, the third (the odd one) is zero-force.

Spotting these first removes members from the analysis and speeds everything up. They are not useless — they brace against buckling and carry other load patterns.

Sign Convention and Tips

Assume every unknown member is in tension (force pulling away from the joint). A negative answer then means compression.
Resolve along the actual member geometry — get the direction cosines right.
Check your work: global equilibrium of the whole truss must hold.

Common Mistakes

Starting at a joint with three or more unknowns (cannot be solved directly).
Cutting a section through more than three unknown members.
Mislabelling tension/compression by ignoring the sign of the result.