Mechanical systems: statics, strength of materials and mechanisms
Here the exam asks for free-body diagrams, support reactions, stress calculations in bars and beams, and the kinematics of mechanisms. It is the block where a sign error or a missing reaction costs the most marks, because the whole exercise follows from it.
Statics
- Equilibrium: ΣFₓ = 0, ΣF_y = 0, ΣM = 0 (moments about the point that eliminates the most unknowns).
- Supports: a pin gives two reactions (H, V), a roller gives one (V), a fixed support gives three (H, V, M).
- Internal forces are found by cutting the structure, never by adding them as external loads.
Strength of materials
- Normal stress σ = N/A; shear τ = V/A; bending σ = M·y/I, with W = I/y_max.
- Rectangular section: I = b·h³/12 and W = b·h²/6; circular full section: I = π·d⁴/64.
- Euler buckling: P_cr = π²EI/L_k², with L_k = 0.5L (both ends fixed), L (both pinned), 0.7L (fixed–pinned), 2L (fixed–free).
- Safety factor n = σ_y/σ_working ≥ design value.
Mechanisms
- Degrees of freedom (planar): G = 3(n − 1) − 2j₁ − j₂, where j₁ are one-DOF joints and j₂ two-DOF joints.
- Grashof: a full-rotation crank exists if s + l ≤ p + q (shortest + longest ≤ the other two).
- Gear train: i = z₂/z₁ = n₁/n₂ = M₂/M₁; in a compound train multiply each stage.
Worked example
A simply supported beam of L = 2 m carries a central load F = 6 kN. Each reaction is 3 kN and M_max = F·L/4 = 3000 N·m. For a 40 × 80 mm section: I = 40·80³/12 = 1 706 667 mm⁴, y = 40 mm, W = 42 667 mm³. Then σ = 3 000 000 N·mm / 42 667 mm³ = 70.3 MPa. Against σ_adm = 120 MPa the beam is safe with n = 1.7.
Marks lost
- Not drawing the free-body diagram, or including internal forces in it.
- Forgetting the roller reaction or giving it two components.
- Using h instead of h/2 in bending stresses.
- Mixing N·m and N·mm (1 N·m = 1000 N·mm).
- Applying Grashof without identifying the shortest and longest links, or miscounting joints when computing G.