14a-Problem-6-8 for local buckling-CM#15 For HSS section

Last Updated on September 19, 2026 by Maged kamel

Problem-6-8 for local Buckling- CM #15 for an HSS section.

Description of data.

The following slide image briefly describes the content of this Post. I will solve Example 6-8 from McCormac’s book to determine the available Strength of a given HSS section. This Time, the solution will be based on CM#15, and we will confirm our answer using Table 4-3. We solved the same problem (6-8) in Post 14, but using CM#14.

This is a brief description of sections that are not slender for compression members, from Prof. Segui’s handbook, with slenderness ratios included for each shape.

Detailed descriptions of problems 6-8 on local Buckling.

For solved problems 6-8, determine the axial compressive design Strength, Φc Pn, and the allowable design Strength, Pn/λc, of a 24 ft HSS 14x10x1/4-inch column section; the base of the column is considered fixed. From the first part of Table 1-11, we obtain A = 10.80 in2, h/t = 57.10, and Ix = 310.0 in4.

The upper end is assumed to be pinned. Use fy = 50 ksi instead of 46 ksi, as in the same problem, based on CM#14. The effective length factor k for the column is 0.7, but the recommended Value is 0.8. Refer to Table C-A-7.1 of the specification. Please refer to the slide image below.

This is part 2 of Table 1-11, from which we obtain the values of ry and Iy: ry = 4.14 inches and Iy = 186 inch4 for the HSS section 14x10x1/4 inches.

Check which direction controls the design (X or Y).

The next step is to check which direction controls the design. For the x-direction, we estimate Lex/rx (0.8*24×12/5.35)=26.92.

While in the y-direction, Ley/ry (0.8*24×12/4.14)=55.65. We select the bigger Value; the design is governed by Buckling in the Y-direction.

Find FE, check whether the HSS section is long or inelastic.

Next, check whether the column is inelastic or long; the limiting ratio is 4.71*sqrt(E/Fy) = 4.71*sqrt(29000/50) = 113.43. Since ley/ry equals 55.65, which is less than 113.43, the column is inelastic. For the Euler stress, we have fe = Pi^2*E/(le/ry)^2; the Fe Value equals 92.42 Ksi.

Check whether the HSS section is slender.

To determine whether the HSS section is slender, we check the b/t Value against the limiting Value for the slenderness ratio for a stiffened HSS section.

The limiting Value equals (1.4* sqrt( E/Fy)) = 33.72. The b/t = 39.90 from Table 1-11, while h/t = 57.10. The section is slender since b/t & h/t are greater than 33.72. Please refer to the next slide for more details.

The following slide image shows the relationships between L/r and critical stress values, along with the curves used to estimate the critical stress.

The following slide image shows the hatched areas representing the effective areas for slender HSS and W sections; the modified b and h are to be estimated to obtain the effective Area.

How do we estimate the effective B and h?

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 td, where t design equals 0.233 inches, h=14 inches, and b=10 inches.

The Value of b=10-3*(0.233) =9.301 inch, while h =14-3*(0.233)=13.301. Since the column is inelastic, we estimate fcr as 0.658^(50/92.42)*50 = 39.87 ksi.

Find the modified B Value.

To determine whether we need to consider Be or the same b Value, check the λ Value against λ, Fy, and fcr. Use equations E7.2 and E7.30.

Check whether λr, the ratio from Table B4.1, equals 33.72, against λ, which equals sqrt(Fy/Fcr)=33.76. Since b/t from the Table is> 37.766, we need to adjust B and estimate be. Later, we will also check for h.

We need to estimate the elastic local Buckling stress, Fel, which equals (C2*λ/λ)^2*Fy. The Value of Fel is 68.01 ksi. We can estimate Be using the E7.3 equation. The final Value of be equals 8.974 inches.

Find the modified h Value.

Check the λr Value from Table B4.1 (33.72) against λ = sqrt(Fy/Fcr) = 33.76. Since h/t from the Table > 37.76, we need to adjust h and use equation 7-3.

We need to estimate the elastic local Buckling stress, Fel, which equals (C2*λr/λ2)^2*Fy. The Value of Fel is 33.21 ksi; it differs because the Lambda Value is higher. We can then find He using the E7.3 equation. The final Value of He, or the effective length, equals 9.923 inches.

We can estimate the effective Area of the HSS section as 9.074 in2. The nominal Load, Pn, equals Ae*Fcr = 9.074*39.87 = 361.78 kips. Use the effective Area to estimate the nominal Load.

LRFD Value for Nominal Load.

For LRFD, Φc=0.90,Φc*Pn=0.9*361.78=325.60 kips.

ASD Value for Nominal Load.

For ASD, Pn/Ωc = 361.78/1.67 = 216.63 kips.

Use Table 3.1 to get the factored loads.

We will estimate the L_e for the lex, Buckling length in x; we will divide by rx and multiply by rx, Lecy Equivalent = 19.20 /(5.35/4.14)=14.86 feet, while lcy=19.20 feet, which is bigger, and that is the length that will be used in Table 4-3.

The factored fcr based on the LRFD design is 326 kips, which is very close to our estimate.

The factored fcr based on the ASD design is equal to 216.4 kips; after interpolation, it is very close to our estimate.

In the footnote, you can check the gross Area Ixx and I, and the rx/ry ratio for the given section.

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

If you want to check the previous problem, see this Link 5-3 for local buckling of columns.

The following Post, Link 15, is 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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