Required Moment of Inertia for Deflection
Mediumdeflectionserviceability
A 6 m simply supported steel beam carries a total service uniform load w = 8 kN/m (self-weight already included). E = 200 GPa and the deflection limit is L/360. Use Δmax = 5wL⁴/(384EI). What is the minimum required second moment of area I?
- A40.5×10⁶ mm⁴
- B56.7×10⁶ mm⁴
- C40.5×10⁹ mm⁴
- D27.0×10⁶ mm⁴
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Answer: A. 40.5×10⁶ mm⁴
Equation: I = 5wL⁴/(384E·Δallow). Convert units: L = 6,000 mm, E = 200,000 N/mm², w = 8 N/mm (1 kN/m = 1 N/mm). Δallow = 6,000/360 = 16.667 mm. Substitution: I = 5(8)(6,000⁴)/[384(200,000)(16.667)] = 40.5×10⁶ mm⁴. Wrong: 56.7×10⁶ uses a factored load (1.4w) for a serviceability limit; 40.5×10⁹ leaves E as 200 (GPa not converted to MPa); 27.0×10⁶ uses L/240 instead of L/360.
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