 # 2a-Buckling for columns-part 2.

## Buckling for columns-part 2.

### A Solved problem for the major and minor axis.

The major aspect of buckling for column- part 2 is to define which axis, major or minor will control the buckling. This is the video used for the illustration of the content of this post.

To decide which will control the buckling we need to estimate the value of(KL/r) about both the major and minor axes and select the highest value out of these two values.

The higher value will control the buckling this will imply estimating of stresses for comparison. While if we are dealing with the load we will select the higher(Kl) value, the numerator has the value of π^2 EI for each axis.

The next solved problem for the major and minor axis will guide us by estimating the controlling compressive load.

It is required to estimate the buckling strength for a column, that has different end conditions for both major and minor axis.
Determine the buckling strength for a w section of 12×50, for a column with height =20 ft for the minor axis, the column is pinned at both ends.

The above slide image shows the minor axis bent from both ends on the right side. While for the major axis Buckling is fixed in one end and pinned on the left side

We start solving this problem by using Euler’s formula for Pcr=π^2 EI/(KL)^2 since this is a W section, we need to use table 1-1 page 1-26 to get the inertia value Ix for the major- axis and Iy for inertia value for the minor axis. For the major axis direction, E=29*10^6 psi we have Ix =391 inch4, k value =0.70, since it is fixed-pinned, we will use k=0.80 as recommended by the code.

While for the minor direction, we have Iy=56.30 inch4, k=1.00 since the column is pinned-pinned.

We will substitute for x-direction Pcr at x =π^2 EI/(KL)x^2=(29*10^6)*391/(0.8*20*12)^2*1/1000= 3038 kips.  for y-direction Pcr at y =π^2 EI/(KL)y^2=29*10^6)*56.30/(0.8*20*12)^2*1/1000= 280 kips.  we will select the smaller value of 280 kips.
The minor axis governs the buckling.
Check whether a column will yield before buckling with different lengths =10 feet.

### The same Solved problem for the major and minor axis which governs buckling?

We will solve the same problem, but after changing the height from 20 feet to 10 feet, the stress will be=76.7 Ksi, which cannot be accepted since it is bigger than the yield stress, which means that the column will yield before buckling.

The difference between the braced and unbraced frames.
Columns are not isolated from other surrounding members as can be seen from the shown sketches. There are braced and unbraced frames, with different shapes of chevron, or k brace.

Due to bracing the frame will be having slight or no-lateral movement.

This is the pdf file that is used for the illustration of this post and the previous post.

The next post: Column compressive strength by the general equation.
For a good reference from Prof. T. Bart Quimby, P.E., Ph.D., F.ASCE, refer to this link for public concepts. Scroll to Top
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