Euler Buckling 1

[Bert2]

Hi all,

I have a question regarding the application of the Euler Buckling theory.

Q - Can this theory / equation be applied to a horizontal 'H' beam with a U.D.L? with the flanges in the Horizontal axis, ie; the web will be buckling / taking the compression load. (Boundary conditions fully fixed i assume)

Thanks.

 

[rb1957]

yes,sort of; it's called a beam column ... a beam with large compression axial loads and transverse loads.

Quando Omni Flunkus Moritati

 

[fegenbush]

Depending on your specific shape, loadings, unbraced length, etc. there can be several failure modes of a beam with a uniformly distributed load. These are Yielding, Lateral-Torsional Buckling, Flange Local Buckling, and Tension Flange Yielding. These are outlined in Table User Note F1.1 of the AISC Steel Construction Manual or in AISC 360-10. The steel construction manual also goes into detail on the calculation of these failure modes. Euler buckling does constitute a portion of the calculation.

 

[Bert2]

rb1957 and fegenbush, thanks for your comments.

It still seems a little unclear to me.

the 'H' beam will only have a compressive force on the Web ie; the 'H' beam is in the horizontal plane. U.D.L in the vertical axis

No axial load will be applied (in the horizontal plane),

Specifically i see the problem area in the second moment of area Ixx calc to include or account for the length as the bd^3/12 form of the standard equation.

if that makes sense?

 

[fegenbush]

Bert2,

What you are describing is a fixed-fixed beam. This is covered in the aforementioned reference at great length. The standard AISC 360-10 is available as a free download from

https://www.aisc.org/WorkArea/showcontent.aspx?id=26516

Chapter F is applicable to members subjected to flexure.

 

[rb1957]

"Euler buckling" to me means column ... Pcr = pi^2*EI/L^2.

you've described a beam (with an "H" cross-section) with a UDL ... pretty standard.

are you asking about allowables to the flanges of "H", buckling under the bending stress ? to me looks like a flat plate, 1 side free, 3 sides SS ... yes?

Quando Omni Flunkus Moritati

 

[Bert2]

Consider the beam is fixed in the Y axis ie; fully supported.

Attached sketch for clarity

 

[rb1957]

thx, that clarified it for me ...
consider 1" of span, loaded by 1" of UDL, I = t^3/12, L = H (depth of beam) ... no?

Quando Omni Flunkus Moritati

 

[Bert2]

rb1972

thanks ive calculated a critical load of 89.15kN for a 20mm section / thickness with a total length of 3840mm and E as 200GPa does that sound reasonable ?

Thanks for the help!

 

[Bert2]

sorry forgot to add my UDL = 28.9kN/m

 

[rb1957]

your beam is 4m deep ?! i get something like that ... 89kN/m

Quando Omni Flunkus Moritati

 

[JStephen]

Your sketch assumes the top flange isn't moving sideways. If it is restrained in some way, that's a reasonable assumption. If it is not, the whole top flange would move sideways.

You'll need to look into the proper K-factor to use given the end restraints.

If it's a rolled beam, you have corner radii that would act to shorten the effective length. I don't know how you'd figure the amount.

But yes, the situation can be handled as a column.

 

[Tmoose]

providing the picture changed completely what I was envisioning from the first post

 

[boyze]

this might help out

online Euler buckling calculator

Have Fun!

James A. Pike

http://www.xl4sim.com

http://www.erieztechnologies.com