A Beginner's Guide to the Steel Construction Manual, 13th ed. (old) Chapter 7 - Concentrically Loaded Compression Members © 2006, 2008 T. Bartlett Quimby |
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Section 7.8 Example Problem 7.1 Last Revised: 11/04/2014 The example problems presented in this section have a spreadsheet solution. You will need this file to follow along with the presented solutions. You can click on the following link to get the file: Chapter 7: Excel Spreadsheet Solutions In addition to the spreadsheet solution this problem also has a downloadable hand solution: BGSCMExample7_1.pdf Given: The following members are being considered for use as a concentrically loaded column. The column has effective lengths, KL, of 15 ft in one direction and 6 ft in the orthogonal direction. The load consists of 1 part Dead Load, 1.5 parts Live Load and 2 parts Snow Load. a) W10x19 Wanted: Determine the capacity of each section for each condition. Consider both ASD and LRFD, expressing the results in comparable terms. Solution: The intent of this problem is to demonstrate how to compute the axial capacity of several different steel shapes. The preferred steel is used in each case. (See SCM Table 2-3, SCM pg 2-39.) The problem statement does not state the orientation of the cross sections relative to the two principle effective lengths. It is wise to orient the members so that the strong axis is associated with the longest effective length. This will cause you to get a lower slenderness, KL/r, for the member. This was done in the solutions provided for this problem. Before solving for the capacity of the individual sections, the load combination equations are used to develop composite load factors (CLF). The CLFs can be applied to the results to get results that are comparable between LRFD and ASD. The provided solutions indicate the controlling load cases. These were determined separately. You should be able to arrive at the same conclusion regarding the controlling load combinations. Note that the ASD CLF is not 1.0. This illustrates the need to unfactor the ASD equations as well as the LRFD equations. Due to the fact that a large percentage of the applied load is less predictable than dead load, ASD will yield more capacity than LRFD in each case. You will see this as you examine the solutions. The steps taken for each problem are pretty much the same in each case:
Note that in part (b) that the HSS section is slender. This necessitates that Q be computed. In this case, since there are no unstiffened elements Qs equals 1. In computing Qa we find that the computed be is greater than the actual b. The result is that the section area is not reduced and Qa is also 1. The product of Qs and Qa is Q, which also ends up to be 1. This means that the slenderness of the cross section did not affect the results of the problem. You may also notice that we used the tabulated value for h/t for the b/t ratio in SCM Table B4.1 case 12. This is because all the section is uniformly compressed and we are to use the largest width/thickness ratio. For part (c) we use SCM Table B4.1 case 15 to determine section slenderness. In this case the section is not slender. As a round section has no principle axis, the radius of gyration, r, is the same for each general slenderness parameter. While the slenderness was computed for both directions, in actuality, only the longest needed to be computed since it is obvious that the shorter direction will not control. To get a better feel for the behavior of the equations, you should set up a spreadsheet similar to the one provided to experiment with altering effective lengths, load combinations, and sections. <<< Previous Section <<< >>> Next Section >>>
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