Last Updated on September 17, 2026 by Maged kamel
Introduction to Buckling for columns-part 2.
A first Solved problem for the critical Load Value.
To decide which will control the Buckling, we need to estimate (KL/r) for both the major and minor axes and select the higher of the two.
The higher Value of (Kl/r) will control the Buckling, implying that the estimated stresses are used for comparison. For the Load, we select the higher (Kl) Value. The numerator is π^2 EI for each axis.
The following item is the first solved problem for the major and minor axes, which will guide us in estimating the controlling compressive Load.
Estimate the buckling strength of a column with different end conditions for both the major and minor axes.
Determine the buckling strength of the W2x50 steel shape, pinned at both ends, with a minor-axis column height of 20 ft.
The slide image above shows the minor axis bent at both ends on the right side. For the major axis, the Buckling is fixed on one end and pinned on the left side.
We solve this problem using Euler’s formula for the critical Load: Pcr = π^2 EI/(KL)^2. Since this is a W section, we use Table 1-1 (pages 1-26) to obtain the Inertia values Ix for the major axis and Iy for the minor axis. For the major axis direction, with E = 29*10^6 psi, we have Ix = 391 in^4 and k = 0.70.
Since it is fixed-pinned, we will use k = 0.80, as recommended by the code. We will substitute for x-direction the critical Load is Pcr at x =π^2 EI/(KL)x^2=π^2*(29*10^6)*391/(0.8*20*12)^2*1/1000= 3035.80 kips.
We have Iy = 56.30 in^4 and k = 1.00 for the minor direction, since the column is pinned-pinned.
For the critical Load in the y-direction, Pcr at y =π^2 EI/(KL)y^2=π^2*29*10^6)*56.30/(0.8*20*12)^2*1/1000= 280 kips. We will select the smaller Value of 280.53 kips. The minor axis governs the Buckling.
Check whether Yielding occurs before Buckling.
For the previous problem, to check whether Yielding occurs before Buckling, we estimate the compression stress. We have Pcr = 280 kips and a W-section Area of 14.60. The stress equals P/A = 280/14.60 = 19.17 ksi, which is less than the yield stress of 50 ksi, indicating that Buckling occurs before Yielding.
Case 2: Yielding occurs before Buckling.
If the column height is reduced to 10 feet, the critical Load is 1119.94 kips; the stress is 1119.94/14.60 = 76.70 ksi, which exceeds the yield stress of 50 ksi; in this case, Yielding occurs before Buckling.
Please refer to Professor Varma’s notes in the next slide image.
A second Solved problem for the major and minor axes.
We will modify the end conditions for the column included in the first solved problem, while keeping the column’s height at 20 feet. The column in the major direction (()x) is pinned at both end. The column in the y-direction is pinned at one end and fixed at the other end.
the critical load is Pcr at x =π^2 EI/(KL)x^2=π^2*(29*10^6)*391/(20*12)^2*1/1000= 1942.9 kips.
Please refer to the next slide image for the estimated Load for the column in the y-direction. The final Pcr Value is 437.12 kips in the y-direction. This Load is chosen because it is less than Pcr at x = 1942.90 kips.
The next slide shows the different types of frame bracing. Here are X and K braces, and bracing can be done using Shear walls.
The difference between braced and unbraced frames. Due to bracing, the frame will have little to no lateral movement.
You can view or download the PDF for this Post and the previousPostt from the following Link.
The next Post: Column compressive strength by the general equation.
For a good A Beginner’s Guide to the Steel Construction Manual, 14th ed. Chapter 7 – Concentrically Loaded Compression Members.
For a good A Beginner’s Guide to the Steel Construction Manual, 15th ed., Chapter 7 – Concentrically Loaded Compression Members.
For a good A Beginner’s Guide to the Steel Construction Manual, 16th ed., Chapter 7 – Concentrically Loaded Compression Members