Engineering Mechanics and Strength of Materials: Equilibrium to Buckling
Statics and equilibrium
A body in equilibrium satisfies ΣF = 0 and ΣM = 0. Draw the free-body diagram first, replace supports by reactions (roller: one normal; pin: two components; fixed: two forces plus a moment), and treat a two-force member as carrying force along its axis.
Stress, strain and beam theory
- Hooke's law σ = Eε, with Poisson's ratio ν = −ε_lateral/ε_axial.
- Bending: M/I = σ/y = E/R; maximum stress at the extreme fibre, σ = My/I.
- Solid circular shaft in torsion: T/J = τ/r = Gθ/L, with J = πd⁴/32.
- Euler buckling: P_cr = π²EI/L_e², with L_e = L, 0.5L, 0.7L and 2L for pinned–pinned, fixed–fixed, fixed–pinned and fixed–free.
- Principal stresses: σ₁,₂ = (σx + σy)/2 ± √[((σx − σy)/2)² + τ²].
Worked example
A simply supported beam of span 4 m carries a UDL of 10 kN/m over its whole length. The section is 100 mm × 200 mm. Find the maximum bending stress. Maximum BM = wL²/8 = 10 × 16/8 = 20 kN·m = 20 000 N·m. I = bh³/12 = 0.1 × 0.2³/12 = 6.667 × 10⁻⁵ m⁴, and y = 0.1 m. σ = My/I = 20 000 × 0.1/(6.667 × 10⁻⁵) = 3.0 × 10⁷ Pa = 30 MPa.
Traps
- Using wL²/8 for a cantilever (it is wL²/2) or wL/4 for a central point load.
- Substituting the diameter instead of the radius into τ = Tr/J.
- Mixing N, mm and MPa; remember 1 MPa = 1 N/mm².
- Forgetting that shear stress is zero at the extreme fibre, where bending stress peaks.