Unbraced Length for Column with Intermittent Loading

[HDStructural]

Hello all,

I have a question regarding unbraced length for a column with intermittent loading. This is for a planar pipe bent (see photo below). The forces in the column increase as they go down due to the lateral load at the top. For the purpose of this discussion, I am not worried about the columns buckling to the left or right, but only in and out of the page. It seems overly conservative to say that the unbraced length for the bottom section with the highest load is 30' with k=2.1. I am familiar with the paper "Elastic Buckling of a Column Under Varying Axial Force". Using this method will give me a reduced overall k value to apply to the column as a whole.

I am analyzing this in STAAD. Do you think it is legitimate to say the unbraced length (into the page) for the bottom section is 7.5' with k=2.1, then 15' with k=2.1 for the next section and so on working my way up? Or should I stick with the method above using an unbraced length of 30' for the whole column with a reduced k value calculated per elastic buckling method?

Thanks

 

[WinelandV]

Are the pipes landing directly on the bent, or is there a pipe bridge spanning between the bents? If there's a pipe bridge, then I'd say the top is braced in/out of the page by the bridge and thus k=1.0. However, this assumes that SOMEWHERE on the layout there's a structure [like a 4-leg tower) that will prevent movement of the in/out direction.

Please note that is a "v" (as in Violin) not a "y".

 

[HDStructural]

There is nothing except for the pipes spanning between the bents. They are not braced in and out of the page. The pipes having sliding connections so that no thrust is applied to the bents

 

[WinelandV]

If there's nothing out of plane, then there's nothing out of plane. L = 30', and k = 2.1.

EDIT: I am not familiar with "Elastic Buckling of a Column Under Varying Axial Force". While it may be applicable, I've never seen industrial bents designed with a reduced k based on a varying axial force.

 

[HDStructural]

The problem that I am having is that the bottom 7.5' section is controlling the design and appears to be overstressed based on L=30' and k=2.1. It seems overly conservative to me to treat the bottom 7.5' of that column the same as the top of the column since they won't have the same tendency to buckle.

 

[JAE]

Out of plane - that lower horizontal isn't even participating in bracing the column...it's just going along for the ride.
There may be some legitimacy to that lower horizontal preventing torsional twist in the column but that doesn't affect a cantilevered column from buckling at the base.

And it appears the base only has perhaps 2 bolts - not really much of a moment connection anyway, right?

What is the intended load path for in-and-out of the page forces?
Maybe the supported pipes have adequate strength and rigidity assuming some fixity between pipe and your frame - but we've always assumed that pipes aren't structural.

 

[HDStructural]

JAE - The bases are fixed going in and out of the page so the bent acts like a cantilevered column in and out of the page (the out of plane load would be wind load). For the unbraced lengths, I am not counting on the horizontals bracing it in and out of the page. My logic is more that the bottom section is only 7.5' from the base so it is seems too conservative to design that bottom section (highest stressed) as if it had an unbraced length of 30'.

 

[ETX]

Try this. A similar question was addressed in the 2010 fourth quarter engineering journal. See attached.

 

[JAE]

The buckling mode doesn't "know" that there's a 7.5 ft. section involved - that's just your view of "sections" along the length.

As you said, "I am not counting on the horizontals bracing it in and out of the page" so there are no sections here.

The physical behavior will be as a single 30 ft. column bending over.

 

[Denial]

Generalising from ETX's 03Jul24@20:50 post above.

Consider a vertical free-standing cantilever of length L loaded with a concentrated downwards force applied at its top.[ ] Euler tells us that this will buckle at
P[sub]cr[/sub] = [π][sup]2[/sup]*EI/(2L)[sup]2[/sup]
If the downwards force is instead applied linearly down the length of the cantilever, beginning with zero force at the top and increasing to a maximum value at the bottom, then it can be shown that the cantilever will buckle when the total load reaches three times this value.


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