Last Updated on March 1, 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 the value of (KL/r) on 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 estimated stresses are used for comparison. If we deal with the load, we will select the higher(Kl) value. The numerator has a value of 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.
Estimating the buckling strength for a column with different end conditions for both the major and minor axes is required.
Determine the buckling strength for the given steel shape W2x50, which is pinned at both ends and has a column height of 20 ft for the minor axis.
The above slide image shows the minor axis bent from 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, estimated as Pcr = π^2 EI/(KL)^2. Since this is a W section, we need to 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, while the W-section area is 14.60. The stress equals P/A=280/14.60=19.17 ksi, which is less than the yield stress of 50 ksi, which indicates that buckling occurs before yielding.
Case 2: yielding occurs before buckling.
If the previous 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 50 ksi, the yield stress; 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 ends. 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 estimate of the critical load for the column at the y- y-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 bracing for frames. There are X and K braces, and bracing can be done using shear walls.
The difference between the braced and unbraced frames. Due to bracing, the frame will have slight or no lateral movement.
The PDF for this post and the previous post can be viewed or downloaded 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.
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