Truss Analysis

A truss is a structure of straight members connected by pins at their ends, its joints, and loaded only at the joints. A pin cannot transmit a moment, and no load acts along a member, so each member is a two-force member. It carries only an axial force N: tension (pulling its joints together, N > 0) or compression (N < 0). Triangles make a truss rigid; that is why bridges and roofs are built from them.

Every joint is a point in equilibrium under the loads on it, the reactions of its supports, and the forces of its members, which act along the members:

A member of length L stretches by δ = NL/(EA), and the joints move so that all members fit together again. The member stress is σ = N/A.

• A pin support stops a joint from moving (two reactions). A roller stops movement normal to its surface only (one reaction): on the floor a vertical reaction, on a wall a horizontal one.
• Each joint gives two equilibrium equations, so a truss with j joints has 2j equations for the m member forces and r reactions. If m + r = 2j and the truss is stable, it is statically determinate: equilibrium alone gives every force. If m + r > 2j it is indeterminate to degree m + r − 2j, and the forces also depend on how the members stretch.
• With m + r < 2j, or with members badly arranged (a panel without a diagonal, or two members in one line at a joint), the truss is a mechanism. The tool detects this, names the joints that can move, and says what to add.
• A zero-force member carries no force under the given loads, for example the third member at an unloaded joint where two members are in line. It is still needed for stability and for other loads.

The method of joints takes the joints one at a time, in an order where each has at most two unknown member forces, and solves equations (1) there. The method of sections cuts the truss through the member of interest and at most two others. Moments about the point where the other two meet give the wanted force from a single equation:

Displacements follow from the unit-load method (virtual work). Apply a unit force at the joint in the wanted direction, find the member forces n it causes, and sum over the members:

An indeterminate truss is solved by the force method. Remove redundant members or reactions until it is determinate, and compute the gaps the loads would open there. Then choose the redundant forces so that the gaps close (compatibility), using the same sums as equation (3).

The step-by-step solution below the simulation works the current truss through these methods, with a free-body diagram for each joint and section. Each probe gets its own tab: a member-force probe by the method of joints or of sections, and a displacement probe by the unit-load method. An indeterminate truss is solved by the force method up to two redundants; beyond that, the stiffness method is summarised.

The scene uses the direct stiffness method. Each member is a bar element with stiffness EA/L along its axis, two displacements per joint. The supports fix some of them, and the system K·d = F is solved for the joint displacements. Member forces follow from how much each member stretches, and the reactions from equilibrium at the supported joints. For a pin-jointed truss loaded at its joints this is exact. A singular stiffness matrix means a mechanism. Its null vector shows which joints can move.

Automated tests compare the results with textbook trusses: a simple triangle, with forces and a unit-load deflection worked by hand, and the classic three-bar indeterminate truss N = P/(1 + 2cos³α). They also check equilibrium at every joint and the energy balance ∑N²L/(2EA) = ½∑F·d for hundreds of random trusses, and that mechanisms are detected.

• Plane truss, frictionless pins, loads only at the joints; the members' own weight is neglected.
• Linear-elastic material, small displacements, the same E and A for every member.
• Buckling of compression members is not checked: a slender member in compression can buckle well before it reaches the yield stress.
• Supports are rigid.