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Chapter 1

Q. 1.2

A two-story office building has interior columns that are spaced 22 ft apart in two perpendicular directions. If the (flat) roof loading is 20{lb}/{ft^{2} } ,determine the reduced live load supported by a typical interior column located at ground level.

Step-by-Step

Verified Solution

As shown in Fig. 1–9, each interior column has a tributary area or effective loaded area of A_{T}=\left(22 ft\right) \left(22 ft\right) =48 ft^{2} A ground-floor column therefore supports a roof live load of

F_{R}= \left(20 {lb}/{ft^{2} }\right) \left(484 ft^{2} \right) =9680 lb =9.68 k

 

This load cannot be reduced, since it is not a floor load. For the second floor, the live load is taken from Table 1–4:

TABLE 1-4   .  Minimum Live Loads
             Live Load               Live Load
Occupancy or Use psf {kN}/{m^{2}} Occupancy or Use psf {kN}/{m^{2}}
Assembly areas and theaters Residential
   Fixed seats 60 2.87    Dwellings (one- and two-family) 40 1.92
   Movable seats 100 4.79    Hotels and multifamily houses
   Dance halls and ballrooms 100 4.79       Private rooms and corridors 40 1.92
   Garages (passenger cars only) 50 2.40       Public rooms and corridors 100 4.79
Office buildings Schools
   Lobbies 100 4.79    Classrooms 40 1.92
   Offices 50 2.40    Corridors above first floor 80 3.83
Storage warehouse
   Light 125 6.00
   Heavy 250 11.97

L_{0}=50 {lb}/{ft^{2} } Since  K_{LL}=4 ,then 4A_{T}=4\left(484 ft^{2} \right) =1936 ft^{2} and 1936 ft^{2} \gt 400 ft^{2} , the live load can be reduced using Eq. 1–1. Thus,

L=L_{0} \left(0.25+\frac{15}{\sqrt{K_{LL}A_{T} } } \right) (FPS units)

L=L_{0} \left(0.25+\frac{4.57}{\sqrt{K_{LL}A_{T} } } \right) (SI units)

L=50 \left(0.25+\frac{15}{\sqrt{1936 } } \right)=29.55 {lb}/{ft^{2}}

The load reduction here is \left(29.55/50\right) 100 \%=59.1 \% \gt 50\% O.K.

Therefore

F_{F}=\left(29.55{lb}/{ft^{2}}\right) \left(484 ft^{2} \right)=14 300 lb=14.3k

The total live load supported by the ground-floor column is thus

F=F_{R}+F_{F}=9.68k+14.3k=24.0 k