14-Solved problem 6-8 for local Buckling of columns-CM#14

Last Updated on September 19, 2026 by Maged kamel

Solved problem 6-8 for local buckling of columns-CM#14.

Description of data.

The following slide image briefly describes the content of this Post.

The previous slide shows shapes for various Q values, the Euler graph, and the limiting vertical line for short and long columns. The formula for local Buckling of columns is included.

Our F_cr Value, when Kl/r > 4.71*sqrt(E/Q*Fy), is Fcr = 0.877*Fe. Our F_crValuee,  when Kl/r>4.71*sqrt(E/Q*Fy), Fcr=0.877* Fe.

The yellow curve on the left side of the vertical line, for the case of kl/r<=4.71*sqrt(E/Q*Fy), when Qfy/fcr<=2.25, then fcr = Q (0.658^QFy/Fe)*QFy, and the different Value of Qs, from A Beginner’s Guide to the Steel Construction Manual, and the shapes of the graph accordingly.

Next, we cover Chapter E: Design of Members in Compression. Part E3 covers short and long columns, and Table E1.1 notes the selection Table for applying Chapter E sections.

For W-sections without slender elements, in the limit states in Chapters E3 and E4, Fb stands for flexural Buckling and Tb for torsional Buckling.

For slender elements, when λf > λr (where λf = Flange b/t and λw = Web b/t >), E7 controls the design and relevant parameters. All elements are for flexural and torsional Buckling.

The third item in the following Table is the tube section (E3), without the slender section. When λ < λr, it is a stiffened section; when λ > λr, it is slender.

Then, E7 will address slender elements for the limit states of Buckling and flexural Buckling.

This is part E7: members with slender elements; the coefficient for the short and long columns for items E7-1 to E7-3. The Q = net reduction factor accounts for slender compression elements: Q = Qs*Qa, when λ > λr for the Flange and Web.

For a cross-section composed only of unstiffened slender elements, since Q = Qs*Qa and Qa for unstiffened elements is 1, then Q = Qs. For an angle, the section is unstiffened.

For cross-sections composed only of stiffened elements, Q = Qs*Qa; for a tube, Q = Qa.
For the general condition Q = Qa*Qs, sections comprise multiple unstiffened slender elements.

It is conservative to use smaller Qs from the more slender elements to determine the member Strength on the next slide; for slender stiffened elements, the parameters for Qa and be are the reduced effective breadth.

The following slide image contains Table B4.1A for compression members for unstiffened elements.

The following slide image contains Table B4.1A for compression members for stiffened elements.

This slide briefly describes non-slender compression member sections from Prof. Segui’s handbook, including slenderness ratios for each shape.

The next slide shows the data for section E7, along with the different values of the critical stress Fcr.

What are the items Qs, Qa, and Qs?

How can we find Qa and be values?

Detailed descriptions of problems 6-Bu Buckling.

For solved problem 6-8, determine the axial compression Strength, Φc Pn, allowable Strength,/λc, of a 24 ft 14x10x114x10x1/4-inch column; the base is considered fixed.

The upper end is assumed to be pinned. Fy=46 ksi. The effective length of the column is 0.7, but the reduced Value is 0.8. Refer to Table C-A-7.1 of the specification.

Properties of the HSS section are from Table 1-11.

Use Table 11-11, parts 1 and 2, for detailed properties. We will get a design section of 0.233-inch wall seAreaArean, the Area is 10.80 inch2, b/t =39.9, h/t=57.1, Ix=310 inch4, rx=5.35 inch2, Iy=186 inch4, ry=4.14 inch2 from part 2.

This is part 2 of Table 1-11, from which we get ry = 4.14 inches and Iy = 186 inch4.

Check whether the HSS section is long or inelastic.

The next step is to check whether the column is inelastic or long; the limiting ratio is 4.7) sqrt(E/fy )= 4.71*sqr t (29000/46)=118.26.

To convert this Value to K*L/r in the y-direction, Kl/r at y = 0.8*24*12/4.14 = 55.652. This is the maximum Value as compared to Kl/r at y since rx> ry.

Ty = y = kl/r at y is 55.652 compared to 118.26. Since kL/r at y < 118.26, the column is short; proceed to find Strength, which is 92.42 ksi, as shown in the figure.

Find Fe and check whether the HSS section is slender or non-slender.

We estimate the Euler stress based on the given section. To determine whether the HSS section is slender, we compare the B/t Value with the limiting Value for the slenderness ratio of a stiffened HSS section. The section is slender since b/t is greater than 35.15. Please refer to the slide image for more details.

Based on B4.1b for stiffened elements, it is recommended to calculate a modified b=b-3*td and a modified h equal to h- 3td, where t design equals 0.233 inches, h=14 inches, and b=10 inches.

Estimate Be and he for the HSS section.

The Value of b=10-3*(0.233) =9.301 inch, while h =14-3*(0.233)=13.301. We estimate be and he as follows: be = 1.921*ts*sqrt(E/Fy)*(1-0.38/b/t*sqrt(E/Fy)).

Please see the following slide for more details. Use Value b for the smaller HSS section width.

Please see the following slide for more details on the HSS section’s larger width. The deducted areas are shown.

Estimate the reduction factor Q.

Since the column is short, divide the effective Area by the gross Area to get Q.

Estimate the critical stress for the HSS section. For the Fcr calculation, Fcr =(Q*fy)*0.658^(Qfy/fe), since Q=0.7972, fcr (0.7976*46)*(0.658)^(0.7976*46/92.49), Our fe=92.42 ksi.

Estimate the factored design Strength values.

Then, the nominal load Pn=Ag*fcr=10.8*Pn = Ag*fcr = 10.8*31.05 = 335.45strength LRFD, Φc=0.90, ΦcPn=0.9*335.45=301.9 kips.
For the ASD, Pn/Ωc = 335.30/1.67 = 200.90 kips.

This solved problem discussed the idea that a section is reduced when a hollow section is treated as a stiffened element due to Buckling. As a result, only the effective Area functions; the effective Area can be estimated and called Aef.
Estimate Qa as Aef/Ag. Then, Value his Q Value to the graph to get Fcr.

This links to a newly added Post 14a: Problem-6-8 for local buckling-CM#15 for HSS section.

This is the PDF file for this Post that can be viewed or downloaded from the following button.

To view the previous Post, 13, refer to: A solved problem 5-3 for local buckling of columns.
The following Post, Post 15, Links to A solved problem 6-19-4.

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.

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