12- Solved Problem 5-2: Local Buckling of Columns.

Last Updated on September 19, 2026 by Maged kamel

Solved Problem 5-2 for local Buckling of columns.

This is a brief description of the content of Post 12- compression.

 Detailed data for solved Problem 5-2 on local Buckling of columns.

A solved problem for local Buckling 5-2, from Prof. Jack McCormac’s handbook.

For the column shown in Figure 5.8, use 50 ksi steel.
a) Using the column critical stress values in Table 4-22 of the Manual, determine the LRFD design Strength, ΦPn,, and the ASD allowable Strength P/Ωc, for the column shown in Figure 5.8.
b) Repeat the problem by using Table 4-1 of the Manual.
c) Solve by using the AISC equation E3.

1-What is the K Value for the column?

Referring to the given section of W12x72, the Height is 15 feet, fixed at the bottom and hinged at the top, so kx = ky = 0.7. However, the recommended values for kx and ky are taken as 0.80.

Using Table 1-1 part 1 for the section properties, the Flange width is bf = 12 inches, the Flange thickness is 0.67, the Web thickness is 0.43, and the overall Height is d = 12.3 inches. There is no C note in the section, which means that the section is non-slender, but we will double-check the slenderness later.

2-Check the bf/2tf, h/tw, rx, and ry for the column.

Using Table 1-1, part 2, for the section properties, bf/2tf = 8.99, hw/tw = 22.60, the radius of Gyration about the x-direction, rx = 5.31 inches, and the radius of Gyration about the Y-direction, ry = 3.04 inches.

We have to determine whether the column is elastic or inelastic. We estimate the controlling lambda=4.71*sqrt(E/Fy)=4.71*sqrt(29000/50)=113.19. Estimate Kl/rx and Kl/ry and check which Value is bigger: Kl/rx = 27.1212. At the same Time, Kl/ry = 47.36; the design is controlled by Buckling in the Y direction, and the column is short.

Please refer to the slide image that shows the relation between Kl/r and the ratio of fcr/Fy. Fcr/Fy will be >0.39 but less than 1.

3-Check whether the column is slender.

Check the ratio b/t should be <=0.56*sqrt(E/Fy),The limiting Kl/r =  0.56 *sqrt (E/fy)=0.56 *sqrt (29000/50) = 13.48. The b/t equals 8.99. The Flange is not slender. There is another check for h/tw that should be <= 1.496*sqrt(E/Fy), which is 35.88.

The h/tw from the Table is 22.60, which is less than 35.88; therefore, the whole section is non-slender.

Qs is 1.0, based on CM#14, when b/t is less than 0.56*sqrt(E/Fy). Please refer to the following slide image for more details.

4-Using Table 4-22 for the available critical stress for compression members.

Refer to the Table for kl/r; for Fy = 50 ksi, our kl/r about the Y-axis is 47.37. This Value is within the 47-48 range.
The Table will give Fcr corresponding to both the LRFD and ASD. For LRFD, interpolation gives a factored stress between 38.3 ksi and 38.0 ksi.
We get 38.189 ksi for the LRFD Value solved in problem 5-2 for local Buckling.

To estimate Φc* Pn=Φc*Fcr*Ag=38.189*21.10=806 kips.

For ASD, refer to the Table for kl/r; for Fy = 50 ksi, our kl/r about the Y-axis is 47.37. This Value is within the 47-48 range.
The Table will give Fcr corresponding to both LRFD and ASD. For LRFD, the factored stress is between 25.50 ksi and 25.30 ksi by interpolation.
We get Pn/ω = 536.0 kips.

If we use Table 4-1 for solved problem 5-2 for local Buckling.

Then, for part b, for the solved problem 5-2, using Table 4-1, as part b for the solved problem, the main difference between the two tables is that Table 4-1 uses the coefficient Kl, not Kl/r as used by Table 4-22, previously our K*l/r=47.37, while our kl= 0.8* 15=12 feet. From kl = 12 ft, draw a line that intersects a vertical line passing through W12x72.

The intersection of the two lines yields 806 kips for the LRFD design. and 536 kips for the ASD design, the same values obtained from Table 4-22, item a; these values match.

If using the general provision for the available Strength.

In the third step, you must estimate the values using the general equation, which is part c of solved problem 5-2. Please find the related equations for the estimation of Fcr values

For the general equation, for the short column, we use the curve on the left side of the 113.43 line.The column is short, Fe=Pi^2E/(kl/r)^2=(Pi)^2(29000)/(47.37)^2, the FE Value equals 127.553 ksi.

Fcr=0.658^(50/127.6557), all multiplied by(50), fcr=42.434 ksi, to be multiplied by phi=0.9 by area=21.1 inch2. We get *Pn = 806 kips for LRFD.

In ASD, for Ω=1.67, Pn/ω = 42.434*(21.1)/1.67 = 536.14 kips, which matches the values in Table 4-22 and Table 4-1.

You can view and download the PDF of this Post from the next document.

The next Post, Post 13, solved problem 5-3 for local Buckling.

For a good A 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.