Statics

Statics is the study of bodies at rest under balanced forces — the foundation of every structure. Learn loads on structures (dead, live, wind, snow, seismic), idealisation, free-body diagrams, supports, internal forces and a systematic method for solving statics problems.

Buildings, bridges, towers, retaining walls and dams are designed to stay still. Statics is how civil engineers find the forces acting on a structure, how those forces travel through beams and columns, and what the foundations must resist.

Every structural calculation starts with statics: idealise the structure, estimate the loads, draw the free-body diagram, and apply the equations of equilibrium.

An analogy

A building is a relay race for loads: the slab passes load to beams, beams to columns, columns to foundations, foundations to the soil. Statics tracks the baton at every hand-off.

Loads on structures

Civil structures carry several kinds of load, each estimated from codes.

Idealisation and free-body diagrams

Real structures are simplified: beams become lines, connections become pins or fixed joints, soil becomes a spring or rigid support. Isolate the body, show every load and reaction, then write equilibrium equations. The quality of the idealisation decides the quality of the answer.

Load paths and tributary areas

A slab spanning between beams sends each beam the load from half the slab on either side (its tributary width). Beams pass reactions to columns; each column collects the load of its tributary floor area on every storey above. Following the load path is how foundations get sized.

Determinacy and stability

A plane structure is statically determinate if the equations of equilibrium alone give all reactions (3 unknowns for a single body). More unknowns → indeterminate (needs deformation analysis); fewer → a mechanism that can move. Structures must be both stable and adequately restrained.

Limit-state design and load combinations

Modern codes (Eurocodes, ASCE 7/ACI/AISC, IS 456/800) check ultimate limit states (collapse) with factored loads such as 1.35G + 1.5Q, and serviceability limit states (deflection, cracking, vibration) with unfactored loads. Combination factors ψ reduce secondary variable loads when several act together.

Formulas

Worked example: Floor beam design load

Given: Slab dead load 3 kN/m², imposed 2 kN/m²; Beam spacing (tributary width) 3 m, span 6 m

Find: Ultimate line load, reactions and maximum moment.

Answer: w ≈ 21.2 kN/m, R ≈ 63.5 kN, M ≈ 95 kN·m

Worked example: Ladder against a wall

Given: 5 m ladder at 60°, weight 200 N (mid-length); Person 700 N at 4 m up the ladder; Smooth wall

Find: Wall reaction and friction coefficient needed at the floor.

Answer: N_w ≈ 381 N; μ ≥ 0.42

Common mistakes

Applications

Key takeaways

Going further