A truss is a structure made entirely of two-force members connected at pin joints, loaded only at those joints. Think of bridge frames, roof structures, and crane booms. Because each member carries only axial force (pure tension or compression, no bending), trusses are extraordinarily efficient — they do maximum work with minimum material.

Key assumptions that make truss analysis tractable:

• All members are straight and connected by frictionless pins.
• Loads and reactions act only at the joints.
• Each member is a two-force member — it can only be in tension (being pulled apart) or compression (being squeezed).

The Method of Joints works by isolating each joint as a free body and applying equilibrium: ΣFx = 0 and ΣFy = 0. Since each joint gives you two equations, you can solve for two unknowns at a time. The strategy is to start at a joint with at most two unknown member forces.

Worked example: Consider a simple triangular truss — a horizontal bottom chord (member AB, 4 m), a vertical right chord (member BC, 3 m), and a diagonal (member AC, 5 m). A pin support at A, a roller at B, and a 10 kN downward load at C.

First, solve reactions. Taking moments about A: R_B × 4 = 10 × 4, so R_B = 10 kN (upward). Wait — the load at C is directly above B only if BC is vertical. Let's place it properly: A at origin, B at (4,0), C at (4,3). Load at C is 10 kN downward. Moment about A: R_B × 4 = 10 × 4, giving R_B = 10 kN upward. Vertical equilibrium: R_Ay + 10 - 10 = 0, so R_Ay = 0. Horizontal equilibrium: R_Ax = 0.

Now analyze joint B (two unknowns: F_AB and F_BC). R_B = 10 kN up. ΣFy = 0: F_BC + 10 = 0, so F_BC = -10 kN (compression, pushing down on B). ΣFx = 0: F_AB = 0 kN. At joint C: F_BC pushes up with 10 kN, load is 10 kN down, and member AC pulls. ΣFy: 10 - 10 + F_AC × (3/5) = 0, so F_AC = 0. In this configuration, the diagonal is a zero-force member — a real and important result. Zero-force members exist for stability under alternate load cases.

Rule of thumb for stability: A truss with j joints needs at least m = 2j - 3 members to be statically determinate. Fewer means it's a mechanism; more means it's statically indeterminate. For our 3-joint truss: m = 2(3) - 3 = 3 members — checks out.

Practical application: Roof trusses (Pratt, Howe, Warren patterns) are designed so that longer diagonal members are in tension (cheaper, thinner steel works in tension) and shorter verticals take compression. Understanding which members see tension vs. compression directly determines material choices and connection design.