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Live Load Reduction for a Floor Member

Mediumlive loadreduction
An eligible member supports one floor with tributary area AT = 400 ft², KLL = 4 and unreduced live load L₀ = 50 psf. Reduction is permitted. Use L = L₀[0.25 + 15/√(KLL·AT)] with AT in ft², but L must not be less than 0.5L₀. Find the reduced live load.
  1. A31.25 psf
  2. B50.00 psf
  3. C25.00 psf
  4. D74.0 psf
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Answer: A. 31.25 psf

Equation: L = L₀[0.25 + 15/√(KLL·AT)], L ≥ 0.5L₀. Inputs: KLL·AT = 4(400) = 1,600 ft², √ = 40. Substitution: L = 50(0.25 + 15/40) = 50(0.625) = 31.25 psf. The floor 0.5L₀ = 25 psf is lower, so 31.25 psf governs. Wrong: 50 psf omits KLL (√400 = 20 gives 0.25 + 0.75 = 1.0); 25 psf takes the lower bound even though the formula result is higher; 74.0 psf puts AT in m² (37.16 m²) into a USCS expression.
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