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Alchen [17]
3 years ago
9

A two-story hotel has interior columns for the rooms that are spaced 6 m apart in two perpendicular directions. Determine the re

duced live load supported by a typical interior column on the first floor under the public rooms.
Engineering
1 answer:
Mama L [17]3 years ago
5 0

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

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