BGASCE7-10 Section 5.8.1
Example Problem 5.1
Last Revised: 11/04/2014
Given: The office building shown in Figure 5.8.1.
Wanted: Compute the total Floor Live load applied to the basement level column at the intersection of grids B & 2.
Solution: For this column, the influence area at each floor level equals the entire floor level because the column supports two girders which, between them, support all the floor joists in the level.
Per level: AI = (50')(36') - (12')(20') = 1,560 sqft
The column in question supports three levels, giving a total influence area, AI, of 4,680 sqft. This value can be inserted in the place of KLLAT in ASCE 7-10 equation 4.7-1.
From ASCE 7-10 Table 4-1, the basic live load for an office building is 50 psf (to simplify the problemwe're assuming no corridors, lobbies, etc.)
The reduced live load is computed as:
L = (50 psf)[0.25 + 15/sqrt(4,680 sqft)] = (50 psf) (0.469) = 23.5 psf
Note that ASCE 7-10 4.7 limits the reduction to being greater than 0.40 in the case, so our computed value of 0.469 is what we use.
While we don't need the actual tributary area to compute the reduction factor, we do need the tributary area to complete the problem, so we do it now. The area tributary to the column must be computed, partially, by analytical means since the girder on grid 2 between grids A and C is not uniformly loaded. Figure 5.8.1.1 shows the setup for the analysis. If the floor is subjected to a unit load per unit area, then the FBD for the member is in terms of area and the 'reactions' represent the tributary areas. We leave it to you to verify the support 'reactions' using statics. So, the area on a single floor which is tributary to column B2 is 420 sqft. Since the column in question supports three levels, the total tributary area, AT, is 1,260 sqft.
Figure 5.8.1.1
Analytical Determination of Column Tributary Areas at One Level - Grid 2
The total Floor Live Load on the column now equals the reduced live load times the tributary area:
ANSWER: PL = (23.5 psf)(1,260 sqft) = 29.6 kips
Notice that we did not use the KLL given in the Standard. This is because there is not a case in the Standard which matches our case. Instead we computed and used the Influence Area, AI, directly. For interests reasons only, we can compute the KLL which is applicable here:
KLL = AI / AT = 4,680 sqft / 1,260 sqft = 3.71