5-A Solved Problem 4-9 for available compressive strength.

Last Updated on September 17, 2026 by Maged kamel

A Solved Problem 4-9 for available compressive strength.

This is a summary of the content of Post 5- compression.

Solved Problem 4-9 for critical stress for a given W-section.

We are continuing with the strength-solved problems for compression members and how to use tables to obtain LRFD and ASD values.

Now, we have a new solved problem: problem 4-9 from Prof. O. Segui’s handbook. We have a W12x58, 24 feet long, pinned at both ends and braced in the weak direction at the third point. In the x-direction, the column is braced at the top and bottom; therefore, k = 1. 

The y-direction is braced at the third point, which means bracing for one-third of the column for each 8-foot length.

We have two points. This solved problem makes k*L/r in the x-direction the main criterion for evaluating the buckling Load, and the x-direction slenderness ratio sometimes controls strength by controlling the bracing in the column’s y-direction.  

We will show the effective lengths (KL)x and (KL)y for the x- and y-directions. Please refer to the next slide for more information and the rx and ry values for the W section. 

What are the different ways to find the available compressive strength?

For evaluating available compressive strength, we use three methods. First, use the general equation after checking whether the column is long or short, since this yields two different equations for Fcr estimation, where Fcr is the critical stress.

The second way is to use Table 4-1 with the required (Kl)y length, which is the larger of the Kl required in the x-direction and the Kl due to buckling in the y-direction.

The third way is to use Table 4-22, based on CM#14. Use the larger Value of (Kl/r) between the x- and y-directions to get the factored stress, then multiply by the Area. This Table is replaced by Table 4-14 in CM#15. Please refer to the next slide for more information.

Table 4-1 assumes the column will buckle in the Y direction. That direction controls the design, but if (Kl) is greater than Kl at Y, we convert (Kl)x into a factor (Kl/rx/ry) and compare it with (KL)y, then select the larger Value.

To convert from X to Y, consider Euler’s stress: Fe = Pi^2*E/(Kl/r)x^2. This equals the same stress in the Equivalent Y direction: Fe = Pi^2*E/(Kl/r)y^2. Take the square root of both sides, and then we can see that KL/rx = Kl/ry. The Equivalent Kl at y for x is equal to Kl/rx/ry.

The next slide image shows the Expression for (KL)y eq =Kl/rx/ry. fo our example Kl at x- direction equals 24 ft while Kl at Y=8 feet the Kl Equivalent in Y from the x- direction =24/(5.28/2.51), where rx=5.28 inches and ry=2.51 inches. The x-direction then controls the design.

This mini-map shows how to convert the buckling (KL)x about the major axis to a fake Value (Kl)y, which is used to compare with the original (KL)y about the minor axis to determine the controlling buckling factor.

The next slide image shows the values of (Kl/rx) and (Kl/ry), and that (KL/r)x is the bigger Value.

The maximum Value of (Kl/rx), which is 54.55, will be compared to the limiting Value of 4.71*sqrt(E/fy).

The limiting Value for Fy is 50 ksi, and the Modulus of elasticity is 29000 ksi, which equals 113.43.

Our column is not in the elastic region. Hence, we estimate the critical stress, Fcr, using the relation Fcr = 0.658^(Fy/Fe) * Fy. To calculate Fcr, we need to assess the Euler stress based on buckling in the x-direction. The Euler stress is 96.19 ksi, as shown in the next slide image.

 LRFD and ASD values for the critical stress for solved problem 4-9-General equation.

Once we estimate the Euler stress, we will calculate the critical stress, plug in all the data, and obtain Fcr = 40.22 ksi.

The LRFD nominal compressive strength can be estimated by multiplying phi*Fcr by the Area. The Area is 17.00 inch2. The Phi Value is 0.90. We get 615.40 kips for the LRFD design.

The ASD nominal compressive stress can be estimated by multiplying (1/omega) * Fcr * Area. The Area is 17.00 inch2. The (1/omega) Value is (1/1.67). We can get the Value of 410 kips for the ASD design. The factored LRFD & ASD values for Fcr can be computed from the general equation; these values are shown in the next image.

 LRFD and ASD values for Critical stress for solved problem 4- 9: Use Table 4-1.

We can use Table 4-1 from the construction manual 14 and Specification AISC 360-10. The Table will provide the available strength values for both LRFD and ASD designs. The controlling parameter is (Kl) in the x-direction = 11.41 feet, with phi*Pn = 615.18 kips for the LRFD design.

For the ASD design, we obtained Pn/ω = 410 kips via interpolation; please refer to the following slide.

LRFD and ASD values for Critical stress for solved problem 4- 9, Table 4-22.

We can use Table 4-22 from Construction Manual 14 and Specification AISC 360-10. The Table will provide the critical stress values for both LRFD and ASD designs. The controlling parameter is (Kl/r) in the y-direction.

We have estimated the controlling Value of Kl/r to be 54.44. The factored LRFD Value for Fcr can be computed from Table 4-22: phi*Fcr = 36.24 ksi. We will multiply by gross Area: phi*Pn = 616.088 kips.

The factored ASD Value can be computed from Table 4-22: (1/ω)* Fcr = 24.09 ksi. We will multiply by the gross Area:: (1/ω)* Pn = 409.53 kips.

Thanks a lot, and I hope you will like the content of this Post.

You can view or download the PDF for this Post from the following Link.

This is the next Post, Post 6: the Alignment chart for columns.

For a good Beginner’s Guide to the Steel Construction Manual, 14^th ed. Chapter 7 – Concentrically Loaded Compression Members.

For a good A Beginner’s Guide to the Steel Construction Manual, 15^th ed. Chapter 7 – Concentrically Loaded Compression Members.

For a good A Beginner’s Guide to the Steel Construction Manual, 16^th ed. Chapter 7 – Concentrically Loaded Compression Members.