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A Beginner's Guide to the Steel Construction Manual, 14th ed. Chapter 8 - Bending Members © 2006, 2007, 2008, 2011 T. Bartlett Quimby |
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Section 8.8 Example Problem 8.3 Given: A floor system spans 20 ft and supports a dead load of 75 psf and a live load of 50 psf. Wanted: Select a I shaped section to use as a floor joist. Specify the section and spacing that provides the least weight per sqft of floor area. The final joist spacing is to be between 2 ft and 6 ft and should be on a 3" increment. Solution: The approach taken for this problem is to determine the required capacity for shear, flexure and deflection in terms of required capacity per foot width of tributary area. Other approaches could have been taken, however, this one works as well as any.To get the required capacity per foot width, we take the maximum uniform pressure (found by using given floor loads and the ASCE 7 load combination equations) and multiplying the pressure by a unit width (1 ft). Since the internal force equations are linear with respect to load, we are making use of the fact that the uniform load intensity equals the uniform pressure times the tributary area (i.e. beam spacing in this case): w = ps Where
So, if s equals 1 ft then the resulting internal forces are in terms of the force per foot width (ftw). As done in the prior example problems, the limit states for flexure, shear, and deflection are considered. The resulting capacities are combined with the internal force equations found in case 1 of SCM Table 3-23, pg 3-213, and are used to solve for the beam spacing. The shortest spacing computed from the various limit states controls the design. s = Provided Capacity / Req'd Capacity per ftw The deflection calculation is a bit different. Here we solve for the required moment of interia per foot width. We can then compute the spacing based on deflection as: s = Provided Ix / Req'd Ix per ftw Actually, in this example we solved for inverse of the required Ix per foot width so that we can use it as a multiplier instead of a divisor. Finally, the weight per unit area can be found for each beam by dividing the weight per foot length by the spacing of the beams. This is equivalent to "smearing" the beam weight out over its tributary width. The result is the average weight per square foot of floor area attributed to the beams. The lowest value gives the solution with least weight. Note that the flexural capacity computation is easier than in prior problems. These beams have full lateral support, so LTB need not be considered. Two methods for solving the problem have been provided in the spreadsheet. The first is the hunt & peck method. The spreadsheet has two columns (one for the LRFD solution and the other for ASD) where the calculations are performed. Member sections are selected manually until the best solution is found. The second method is the brute force method. In this case a big table is created with each row being a different section and the calculations being performed in successive columns. As the table is very wide, it has been broken into two tables with the right half being placed below the first half. In the brute force method, the calculations were performed for all sections then sorted by weight per unit floor area. Only those sections that meet the spacing criteria were included in the final list. The least weight solution is then easily identifiable as the first section on the list. Examining the results, it is apparent that, for this problem, total load deflection is the controlling limit state. <<< Previous Section <<< >>> Next Section >>>
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