Mechanics of Structures and Strength of Materials
Questions here test equilibrium, support reactions, shear force and bending moment diagrams, and the basic stress, strain and bending formulae applied to a simple beam or bar.
Equilibrium and beams
- Conditions: ΣFx = 0, ΣFy = 0, ΣM = 0. A simply supported beam carries vertical reactions only; a cantilever's fixed end carries a reaction moment as well.
- For a simply supported beam with a central point load W, maximum BM = WL/4 at midspan and maximum shear = W/2 at the supports.
- For a uniformly distributed load w over span L: maximum BM = wL²/8 at midspan, maximum shear = wL/2 at supports.
Stress, strain and bending
- Direct stress σ = P/A, in N/mm² (same as MPa). Strain e = δL/L. Young's modulus E = σ/e.
- Bending equation: M/I = σ/y = E/R. Section modulus Z = I/y, so σ = M/Z; for a rectangle, Z = bd²/6.
Worked example
A simply supported beam of span 4 m carries a UDL of 5 kN/m. Total load = 5 × 4 = 20 kN, so each reaction = 10 kN. Maximum BM at midspan = wL²/8 = 5 × 16 / 8 = 10 kNm. For a 100 mm × 200 mm section, Z = 100 × 200² / 6 = 666,667 mm³. Convert: 10 kNm = 10 × 10⁶ N·mm. Then σ = 10 × 10⁶ / 666,667 = 15 N/mm².
Marks-losing traps
- Substituting kNm directly into M/Z without converting to N·mm (a factor of 10⁶ error).
- Sign convention: sagging is taken positive at midspan, hogging negative over a support.
- Drawing the SFD through the load point instead of stepping at it.