Statics
Statics
The FE Mechanical statics items test equilibrium of particles and rigid bodies, free-body diagrams, trusses, frames and machines, friction, centroids, and moments of inertia. The exam expects you to identify support reactions, reduce distributed loads, and apply equilibrium equations without being distracted by internal forces.
Core rules and concepts
- Equilibrium in 2D: ΣFx = 0, ΣFy = 0, ΣM = 0. These give three independent equations for a rigid body.
- Free-body diagram: isolate the body and show every external force, reaction, weight, and applied moment. A roller has one normal reaction; a pin has two components; a fixed support has two forces and a moment.
- Trusses: method of joints works when a joint has at most two unknowns; method of sections cuts at most three members. Identify zero-force members by inspection.
- Friction: impending motion uses F_s = μ_s N; sliding uses F_k = μ_k N. Friction opposes relative motion or impending motion.
- Centroids: x̄ = ∫x dA/∫dA. For composite areas, treat holes as negative areas.
- Parallel-axis theorem: I = I_c + A d², where d is the distance between parallel axes.
- A distributed load is replaced by its resultant area acting at the load's centroid.
Worked example
A simply supported beam is 6 m long. A uniform load of 4 kN/m acts over the entire length, and a 12 kN point load acts 4 m from the left end. The distributed load resultant is 4 × 6 = 24 kN at 3 m. Sum moments about the left support A: B_y(6) - 24(3) - 12(4) = 0, so B_y = (72 + 48)/6 = 20 kN. Sum vertical forces: A_y + 20 - 24 - 12 = 0, so A_y = 16 kN.
Traps that cost marks
- Placing a distributed load resultant at the wrong location.
- Forgetting a reaction or assigning the wrong support type.
- Making a sign error in the moment equation.
- Misidentifying a zero-force member.
- Reversing the direction of friction.
- Adding a hole's area instead of subtracting it in a composite centroid.