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 [tex]L_{o}[/tex] of the structure is 4790 N/m^2. The live load element factor ([tex]K_{LL}[/tex]) is 4. The reduced live load will be:
L = [tex]L_{o}(0.25 + \frac{4.57}{\sqrt{K_{LL}A_{T}}})[/tex] = [tex]= 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[/tex]
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*[tex]L_{o}[/tex] = 0.4*4790 = 1916 N/m^2
