Statics, Forces and Structural Analysis
Statics analyses bodies in equilibrium: ΣF = 0 and ΣM = 0. Candidates must resolve forces into components, draw free-body diagrams (FBDs), and apply equilibrium equations to beams, trusses, and frames.
Core concepts: resultant of concurrent forces; moment of a force (M = F·d, with perpendicular distance); couple (two equal opposite forces, moment = F·d); support reactions (pin = two components, roller = one perpendicular, fixed = force + moment). For trusses, assume pin joints and loads at joints, then use the method of joints (ΣFx = ΣFy = 0 at each joint) or method of sections. Identify zero-force members by inspection. Distinguish tension (pulling) from compression (pushing).
Traps
Sign conventions must be consistent; forgetting the perpendicular distance in moments is common. Distributed loads must be converted to a single resultant at the centroid. In trusses, members only carry axial force - no bending. Watch for over-constrained or unstable structures (reactions > 3 for 2D or insufficient support).
Worked rule of thumb
For a simply supported beam of length L with a point load P at midspan: reactions are P/2 each; maximum bending moment = PL/4 at midspan. Example: L = 4 m, P = 8 kN → reactions 4 kN; Mmax = 8×4/4 = 8 kN·m. Always draw the FBD first, then sum moments about one support to find the other reaction.