Lateral Torsional Buckling of Thin Beam

[asalisbury]

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.

 

[271828]

Deleted. Sorry, spoke too soon.

 

[phamENG]

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.

 

[HTURKAK]

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.

 

[dold]

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.

 

[EngDM]

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.

 

[Nick6781]

You probably need to worry about end bearings in additional to bending

 

[asalisbury]

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.

 

[rb1957]

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.

 

[BAretired]

Looks like the OP has no idea what he is asking.

 

[rb1957]

well he is an "EIT" but he should've gotten further into the problem on his own.