Trusses, Frames and Virtual Work

A truss is a structure of straight members joined at their ends by pins, loaded only at the joints. Every member is then a two-force member: it is either in tension or compression.

Determinacy

For a plane truss with m members, j joints and r support reactions:

• m + r = 2j: statically determinate (with r = 3, m = 2j − 3).
• m + r > 2j: indeterminate (redundant members).
• m + r < 2j: a mechanism (unstable).

Method of joints

Take one joint at a time, starting where there are at most two unknown member forces, and apply ΣFx = 0 and ΣFy = 0. Assume every unknown force is tensile (pulling away from the joint); a negative answer means compression.

Method of sections

To find the force in one member in the middle of a truss, cut through at most three members including it, and apply the three equilibrium equations to one side. Take moments about the point where the other two cut members meet: the equation then has only the wanted force.

Zero-force members

Spotting them saves time:

1. At a joint with two non-collinear members and no load, both members carry zero force.
2. At a joint with three members, two of them collinear, and no load, the third member carries zero force.

Principle of virtual work

If a body in equilibrium is given a small virtual displacement compatible with its constraints, the total virtual work done by the external forces is zero: Σ F·δs = 0.

It is useful for mechanisms such as levers, linkages and screw jacks: write each displacement in terms of one coordinate and solve one equation, without finding reactions.

Example (lever): effort P at 2 m from the fulcrum lifts a load W at 0.5 m. A small rotation δθ gives P(2 δθ) − W(0.5 δθ) = 0, so P = W/4.

Worked example

A triangular truss ABC has A pinned, B on a roller, AB = 4 m horizontal, and C at the apex 2 m above the midpoint of AB. A vertical load of 20 kN acts at C.

• By symmetry, R_A = R_B = 10 kN.
• At joint A: member AC is at angle α with tan α = 2/2 = 1, so α = 45°. Vertical: F_AC sin 45° + 10 = 0, so F_AC = −14.14 kN (compression).
• Horizontal: F_AB + F_AC cos 45° = 0, so F_AB = +10 kN (tension).

Practice