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## 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.

## 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

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