Question

A two-story hotel has interior columns for the rooms that are spaced 6 m apart in two perpendicular directions. Determine the reduced live load supported by a typical interior column on the first floor under the public rooms.

Answer 1

Answer:

3021.7 N/m^2 or 3.022 kN/m^2

Explanation:

The area of the interior column is equivalent to 6*6 = 36 m^2. The length $L_{o}$ of the structure is 4790 N/m^2. The live load element factor ($K_{LL}$) is 4. The reduced live load will be:

L = $L_{o}(0.25 + \frac{4.57}{\sqrt{K_{LL}A_{T}}})$ = $= 4790(0.25 + \frac{4.57}{\sqrt{4*36}}) = 4790(0.25 + \frac{4.57}{\sqrt{144}}) = 4790(0.25 + \frac{4.57}{12}) = 4790(0.25 + 0.381) = 3021.7$

Therefore, the value of the reduced live load that will be supported by the column is 3021.7 N/m^2 or 3.022 kN/m^2.

This is less than 0.4*$L_{o}$ = 0.4*4790 = 1916 N/m^2