Calculating required stiffening on slotted column

[tankerator]

My client wants me to repurpose an 40'H x 60'D API tank into a knock out drum and I am having trouble getting started on the slotted center column. The column is a 24" pipe (sched. TBD) with three sections of eight vertical slots to allow the gas escape up the column. I assume there would be stiffeners, like flat bars on end, in between these slots but I don't really know how to calculate the size required in order to replace the moment of inertia lost from adding the slots. I can calculate the moment of inertia on simple objects like pipe and bars but my knowledge gets fuzzy when you start cutting holes. Thanks in advance for your help.

 

[Wrenchbender]

If its a question of finding the polar moment (J) of a hollow cylinder in the places where symmetrical slots have been cut out, J should scale linearly with the material that is left in the cross section.

 

[EngineerTex]

I may or may not agree with Bestwrench on this, depending on how long the slots are.

As for a slightly conservative approximation of member buckling of the eight segments of the slotted section, you can check critical buckling by considering 1/8th of the load acting on a member of rectangular cross-section with the thickness as the wall thickness, the width as the length of the arc of the pipe ID and the height as the length of the slot. Consider this approximation as having ends fixed against rotation.

As for global stability, consider the column as four sections with three stiff hinges in the areas of the slots. But that will be pretty tricky because I really don't know how you would do that. You would likely have to perform FEA to determine if adding these slots would leave the entire column prone to buckling.

Will it see any loads besides the dead weight of the column itself? Will it ever be torqued or eccentrically loaded? That would change everything if it ever is.

Again, I like what Bestwrench said, but if your slots are too long, then you get into a problem where you're not really dealing with a column with reduced properties, rather, you're dealing with eight separate members, each carrying 1/8th of the load.


Engineering is not the science behind building. It is the science behind not building.

 

[tankerator]

Thanks for the direction, that gives me something to think about.

To answer your questions on the column, it will support the rafters and thus the roof. Since the column is attached to the roof bracing, I foresee it experiencing little torsion. The slots are estimated to be 150mm x 1530mm (5.9” x 60.25”) at 8 slots per section, 3 sections equally spaced between elevations 12’ and 32’. That leaves about 30” between sections. The total column height is approximately 65’ tall.

EngineerTex, I like the suggestion of treating it like eight smaller sections to check for buckling. That seems intuitively correct to me. I will definitely try that.

Bestwrench, by scaling linearly, I understand you to mean scale J by subtracting the percentage of material that is removed? That also makes sense. I am worried, like EngineerTex said, that the slots may be too long to be considered part of the column. It does steer me in the right direction as far as adding stiffening bars. It seems to me that taking one of the remaining areas and adding a flat bar (thus making a cross section like a T), I can calculate the polar moment of inertia on that and multiply it by 8 to get the total for the set.

Does that seem logical to either of you? Does the radius of gyration play any part? I know that the radius of gyration is used in buckling but is that only for circular cross sections? I guess I am unfamiliar with how the two are used in applications like this.

Thanks again.

 

[EngineerTex]

To determine if this is an Euler or a Johnson column, you will need to consider the radius of gyration and find the critical slenderness ratio for a member of the cross-sectional properties and then compare it to your member. Then you will use either the Euler or Johnson column formula.

Remember that AISC also requires a minimum SF of 1.92 (23/12) for members in compression and everything in this post only considers columns loaded in compression ONLY.

Radius of gyration is calculated as r = SQRT(I/A).

I = second moment of area (use the weak axis obviously for buckling)
A = cross-sectional area

You will use this for your buckling formula as applied to the 1/8th area. For the length (and width) of the slots you are dealing with, I would not go down the road of deducting a percentage of the overall sectional properties. To prove it, you should calculate it both ways and show yourself how much of a difference there is between the two.

On stiffening:

I don't remember where I read it, but a rule of thumb to prevent first-order buckling only requires a restraint that will see approximately 5% of of the axial load applied at the critical buckling area. In other words, it doesn't take much to stiffen a member against first-order buckling. Instead of flat bars, you might have an easier time putting flat-ring-shaped stiffeners in one or three locations at each slot area. Perhaps try 1/4" plate with a 24" ID and 30" OD welded at the three critical locations for first and second mode buckling on each of the three slot locations (3 or 9 rings total for your column). That would probably do a significant amount of reinforcement if you find that the member cannot handle the load in the slotted area. But only if you need it.

Engineering is not the science behind building. It is the science behind not building.