A Beginner's Guide to ASCE 7-10

Chapter 5 - L: Live Loads

© 2012, T. Bartlett Quimby

Overview

Uniformly Distributed and Concetrated Loads
Loads on handrails, guardrails, grab bars, ladders and vehicle barriers
Arrangement of Live Loads
Live Load Reduction
Roof Live Load Reduction
Crane Loads
Example Problems

Homework Problems

References
Report Errors or Make Suggestions

 

BGASCE7-10 Section 5.8.2

Example Problem 5.2

Last Revised: 11/04/2014

Given:  The office building shown in Figure 5.8.1. 

Wanted:  Compute the total Roof Live load applied to the any column at the intersection of grids A & 2.

Solution:  This column supports a graphically identifiable roof area.  The tributary area, AT, is:

AT = (18')(13') = 234 sqft

All columns at this grid intersection support the same amount of roof.

Notice that the roof has two different slopes to it.  For this problem, we will conservatively use the lesser slope in the reduction computation.  If there is an advantage, we could compute a different live load for each slope, but then we'd also need to determine the tributary area associated with each load--an easy task.  As our goal is to illustrate the use of the reduction formulas, it is sufficient to only consider one slope at this time.

From ASCE 7-10 Table 4-1, we find the roof live load, Lo, to be 20 psf

Next, we'll compute R1.  As AT is between 200 sqft and 600 sqft, we'll need to use the linearly interpolation version of the equation:

    R1 = 1.2 - 0.001(234 sqft) = 0.966

Now, we compute R2.  In our case,

F = (8 ft / 20 ft)*12 = 4.8

Since F is between 4 and 12, we use the linear interpolation version of the equation:

R2 = 1.2 - 0.05(4.8) = 0.960

Now that we have the pieces we can compute the reduced unit roof live load:

L = (20 psf)(0.966)(0.960) = (20 psf)(0.927) = 18.5 psf

The resulting column roof live load is now:

ANSWER:  PLR = (18.5 psf) (234 psf) = 4.34 kips