A Beginner's Guide to the Steel Construction Manual, 15^th ed.

Chapter 7 - Concentrically Loaded Compression Members

© 2006, 2008, 2011, 2017 T. Bartlett Quimby

Section 7.8.1

Example Problem 7.1

Last Revised: 04/19/2021

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

Given: The following members are being considered for use as a concentrically loaded column. The column has effective lengths, L_c, 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
b) Rectangular HSS 10x4x1/2
c) Round HSS 7.5x0.500

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 2-4, SCM pg 2-48.)

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:

1. Compute the general slenderness parameters, L_c/r, and identify the controlling value.
2. Compute F_e.
3. Determine the appropriate equation for computing F_cr.
4. Compute F_cr and P_n.
5. Determine if the cross-section is slender or not.
6. If the cross-section is slender, then compute the effective cross-sectional area, A_e, using SCM E7.
7. Compute P_n.
8. Compute P_s,eq using the composite load factors.

You may also notice that we used the tabulated value for h/t for the b/t ratio in SCM Table B4.1a case 6. 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 B4.1a case 9 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. Note that you should create a spreadsheet the automatically computes A_e if the section turns out to be slender. The provided spreadsheet does not do that.

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