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Lateral Torsional Buckling of Thin Beam

asalisbury

Mechanical
Mar 6, 2024
9
I am working on determining if I have any lateral torsional buckling in this beam. I can't really find any equations for this specific scenario so any help or pointers would be appreciated. It's a very thin and long metal beam that is getting loaded on each end with 5,000lbs. It is pressing into a simple support that covers most of the beams span.

1741797557540.png
 
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Don't think I'd really call that beam as it has essentially zero bending stresses in it. So no LTB, but plate buckling is likely. And then just general stability if there's nothing bracing the top edge.
 
Is this a beam on elastic foundation? Will you provide more info. for ( 25 in simple support, restrains at both ends , ..) and review your figures. If the thk. 0.05 in, the plate will buckle with very small loading before LTB.
 
I don't see how 20mil sheet could handle 50 lbs, let alone 5000 lbs? That's just a hair thicker than a 25 gauge drywall stud.
 
Yea running through this with the typical web bearing at beam end equation (not that it would necessarily translate perfectly without a flange) I get a capacity of 225lbs roughly. There is zero chance this is holding 5000lbs.

A typical paperclip is thicker than the beam you are proposing.
 
My force was not right, it is getting distributed over many of these. Each end is getting 19.5lbs applied. We have a set of combs that hold each end in place and resists torsion. This does seem to fall under plate buckling and not beam buckling. 1741807477754.png
 
I would assume that the load is reacted near both ends, rather than distributed. It would be a "nice" shear lag problem to distribute.

I would assume a 30t width effective in compression (this is a standard assumption in my field) so effective width 0.6".you could use the width of your guides

I would assume a 2" long column, you could finesse with the distance between the guides.

This is steel, yes?

So Peuler = pi^2*EI/L^2 = 10*30E6*(0.02*0.6^3/12)/2^2 = 10*30E6*3.6E-4/4 = 27000 lbs ! but this is a stress of 2E6 psi !!

Your load produces a stress of <2ksi ... which should be good by inspection.
 
We started with a 5000# load at each end. Now, because there are "many of these", the load has reduced to 19.5#, so "many" turns out to be about 260. How far apart are the plates? What type of member can apply a load of 5000# uniformly over a span of 260s where 's' is the spacing of plates?

Does the simple support deflect over its 25" span, or is it deemed to be rigid? If rigid, I agree with rb1957 that the load is carried over a very short length at each end of the support, so LTB over a span of 25" is likely not a problem.

A plan and typical sections of the assembly might help our understanding of the problem.
 
So Peuler = pi^2*EI/L^2 = 10*30E6*(0.02*0.6^3/12)/2^2 = 10*30E6*3.6E-4/4 = 27000 lbs ! but this is a stress of 2E6 psi !!
I think you got your b and t backwards for moment of inertia since it will be buckling in the weak axis direction, no? b = 0.6", t = 0.02".

Peuler = 30 lbs, at 2.5ksi.

Either way, I don't think we have an accurate picture of the problem statement. Is this "assembly" like some sort of heat exchanger or something? Very thin fins supported at the ends by some sort of frame? Like the fins on a radiator. And the force is from the frame being pressed around the fins, or vice-versa?
 

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