Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations KootK on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Web only steel beam?

Status
Not open for further replies.

MHSpurs

Structural
Apr 11, 2005
29
Hi

Has anyone had any experience designing flat plate as a beam?

I need to design a tapered flat plate (250mm x12mm wide tapering to 150x 12mm wide) cantilever beam supporting a glass canopy.

The beam is lightly loaded but has no lateral restraint.

I cant seem to find any design guides or codes that cover the topic.

Any ideas?

 
Replies continue below

Recommended for you

DaveAtkins-
I think it would be unconservative to use 1/2 b just because 1/2 of the plate is in compression. On page 5-35 in the 9th Ed. ASD, the d for the stem of a Tee is the full nominal depth, and the full stem is obviously not in compression.
 
I will agree with DaveAtkins on theory. The b/t ratio was developed from plate buckling under uniform compressive with one or both edges supported. The edge support is accomplished through the fact that this cantilever plate is prevented/restraint from buckle above the neutral axis (tension side), therefore the use of b/2 (for cases without axial load) is applicable. One reason for AISC to designate full depth of the stems as "b" is obviously due to the consideration of beam-column elements subject to both bending and compression forces, for which, the neutral axis varies from case to case, or the entire section is under compression. Since the use of plate as beam element is rare, so most tests & AISC do not address it explicitly, rather research on the development of the code provisions, formulas in conjunction with engineering judgement is necessary.
After all said, in practice, I will agree with StructuralEIT in the use of full plate depth as "b" for engineering shortcomings.
 
Structural EIt

Re your earlier point. Zx does not imply full plastic moment, rather full elastic moment.

Add this to the fact that I intend to reduce the allowable stress due to the cross section being slender, i think I am going down the correct road.


i reason it this way:

Lateral torsional buckling generally deals with sections where the top and bottom flange of a member are extremely strong/effiecient relative to their webs.

The reduction in plastic capacity is due to the fact that the web is rather inefficient in stabilising an unrestrained flange (tension compression couple). hence the moment that can be developed is a function of the ability of the web (and general torsional rigidity) to stabilise two elements that are stresseed to multiples of its own capagity.

Similar to box sections, lateral torsional buckling should not apply. Instead, modifications of the allowble stress due to slendernes are more appropriate.
 
MHSpurs-
I am not seeing why you would use an Allowable stress with the PLASTIC section modulus and call it ELASTIC moment. This is not making sense to me. Zx is plastic section mod., Sx is Elastic section mod.
The only way you can count on the plastic section modulus is if you can develop the full plastic moment of the section before the section will buckle. For a WF section the shape factor is about 1.12, hence the increase in allowable stress from 0.6Fy to 0.66Fy (ONLY IF LTB IS NOT GOING TO OCCUR). For a vertical plate like you are talking about, the shape factor is 1.5, so you would be increasing your allowable stress from 0.6Fy to 0.9Fy.
I think it is a mistake to say that you don't have to worry about LTB because there is no flange.
To compare a vertical plate to a box section makes no sense, in my opinion. A box section usually doesn't have a problem with LTB because Sx=Sy. In your case, Sx is much, much greater than Sy. In addition, a box section has significant torsional rigidity, whereas a vertical plate has virtually zero torsional rigidity.
I seem to agree, generally speaking with your reasoning up (and EXCLUDING) the comparison of this vertical plate to a box section.
Additionally, to use Zx with a reduced allowable stress you would have to use an allowable stress of 0.44Fy to get to an equal moment as 0.66Fy*Sx.
Finally, if LTB didn't apply for this situation, there would be no need to discuss this at all and you would just use a traditional M/S<0.66Fy.
 
One last thing to mention - if LTB didn't occur in a situation that you are describing why in the world would you have to check coped beams for web buckling?
The answer is, you MUST check the webs of coped beams for buckling, so why wouldn't you do it for a vertical plate?
Oh yeah, in a cantilever situation it is actually the tension (top) side that is more critical as far as bracing goes.
 
MHSpurs
I also cannot agree with your idea that the flat plate will not be subject to LTB. LTB doesn't apply to box sections only if they are square or 'under-square'. Rectangular sections being about their strong axis can be affected by lateral buckling.
The Australian Steel Structures standard AS4100 accounts for LTB for different section shapes with a 'slenderness reduction factor'. This is calculated using the section capacity, Iy, effective length, elastic modulii E & G, torsion constant and warping constant.

StructuralEIT
It may be different where you are, but in my experience in Australia and the UK; Zx is the elastic section mod. & Sx is the plastic section mod.
 
Lateral torsional buckling generally deals with sections where the top and bottom flange of a member are extremely strong/effiecient relative to their webs.

lateral torsional buckling should not apply.

LTB is amongst other things a product of lateral stiffness and torsional stiffness. A plate has very little of either of these and will most certainly experience lateral torsional buckling. If you dont believe me put a ruler in a vice and load the end.

If there is no lateral support along the top then LTB will govern - plates fall over pretty darn fast.
 
It would seem to me that either local buckling or LTB would govern. I have not taken a class on plates, but it shouldn't be too difficult to derive the equations for each case.

Since half of your section will be in compression, I would think that it would be comparable to a typical flange. I.e. The bottom half of your beam (compression half) cantilevers from the top half, analagous to the half of a flange which cantilevers from the web, if that makes any sense. So maybe your local buckling limit will be something like h/(2*tw) but don't quote me on that.

As far as LTB, the derivation is pretty general - I would say just set Cw (warping torsional constant)= 0 whenever it shows up. By doing that and noting that Iy is going to be much smaller, the moment capacity should be quite small as expected.

Just my two..
 
As was stated earlier by Structural EIT and lkjh345, this is covered by Chapter F11 of AISC360. As you are using Zx for the elastic section modulus, I'll guess that you don't have access to AISC360.

F11 was derived from article 39, page 202 of Strength of Materials by Timoshenko, Part II, Second Edition, and the derivation of F11 is covered in the SSRC Guide to Stability Design Criteria for Metal Structures.

Do not rely on a finite element analysis, as finite elements do not check for buckling, to my knowledge.

If you do not have access to the above, post the length of the beam, and someone (or two or more) here will crank it for you.
 
According to the AISC steel solutions center, plates may be designed by F11 provided the material they are made of is approved for under the AISC.

 
Hi all.

I didn't anticipate that this q would generate so much interest.

Firstly, Zx is the elastic section modulus in the uk (where I am based). I was not aware that it was reversed in the US. (you say tomatoes etc...) Hope that clears that one up.

Secondly, I am assuming that the glass does not offer lateral restraint to the compression zone (cantilever, glass on top).

The span of the cantilever is approx 1000mm.

The loading is 51kg/m from the glass and potentially 100kg/m from snow.

The plate is 250 high x12 thick at its tip reducing to 150x12 at its support.

Partial torsional restraint can be achieved at the support.

By my calculation (SLS state - I'm old school!)

BM = 0.75kNm (sls); Zxx = 45000mm3;
Py of material = 190N/mm2 (sls)
Slenderness of cross section at main moment position = 150/12 =12.5

Ignoring LTB and working on allowable stress:

M/Z = 16.7N/mm2 which is approx 9% of the allowable elastic stress.

If anyone would like to have a go based on LTB I would appreciate it.
 
If the US code addresses LTB for plates, I wuold expect the British one to do so also. I think you should check LTB instead of relying on a low percentage of the elastic moment. There comes a point when anymore simplification just isn't engineering anymore. It's already a tapered plate with no restraint.
 
We have a cantilever with a length:depth ratio of 7, and a total u.d.load of about 150kg; equivalent to a grown man standing at the end; there is no way that it can fail as I see it.
It will almost work bending about its weak axis.
I'd be happy if it was my design.
 
As UcfSE says, you really should check your own steel code for this. If it's mum on this point, try a strength of materials text as cited above.

That said, according to our code, I've got you in the range for inelastic buckling, with a design moment less than 10% of the design strength.
 
Hi MHSpurs

I am sorry for the delay in responding however I was researching the problem a bit further.
Anyway I analysed the beam on an Excel spreadsheet and got the maximum stress to be 5.9N/mm^2 at the built in end and a deflection of 0.469mm at the free end.
My calculation was based on splitting the beam up into 50mm
long sections and calculating the Z and "I" values at those
points.So for the stress I used M/Z where M was the moment
at whatever point on the beam.
For the deflection I converted the UDL to a point load at the end of the beam and used Strain Energy theory with a graphical intergration method to obtain the deflection.
I also found this free calculator in an Excel spreadsheet
for a tapered beam and the stress is in good agreement with the figure I obtained:-Seemingly though my deflection at the free end didn't agree with there's, nor did it agree with the Pro E simulation figure for deflection, which is run in conjunction with the
spreadsheet.
Finally I do have some reservations about this analysis
and these are as follows:-

1. The beam is very short in length compared to its depth
so it puts a question mark about using simple beam theory
for analysis.

2. Laterally the beam as much less stiffness so I think
the points raised by others are valid unless your
absolutely sure its only going to be loaded vertically
through its centroid otherwise you get into asymmetric
bending.

I assumed that the 250mm end was the built in end but looking at your last post it would appear that its the other way round is that correct?

regards

desertfox
 
Hi Again

deflection should read 0.186mm not 0.469mm

regards

desertfox
 
Hi Desertfox and all others who have replied.

Many thanks for your input.

The design went into fabrication today so no turning back now.

I'm satisfied with the design. Its not something that is covered very well in our code. I think I'll drop a note to the Institution.

Thanks again

MH
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor