Twin beam lateral restraints at top flange 1

[george69]

HI all

I have 2-900WB simply supported beams spanning 18m. the beams are 1100mm apart. There is a large concentrated load in the middle. There is top flange K-bracing that tie the beams together. The nodes that for this "truss" are at 1m centres.

They have the same bending moment at midspan and are working near their section capacity.

The AS4100 steel code says that lateral restraints need to be able to take 2.5% of the flange compression force at the section considered. Is anyone able to elaborate on how these forces are applied/considered?

For instance, do I treat the top flange and bracing as a truss with discrete point loads along the length? i.e a truss analysis!!, OR, is the 2.5% loading meant to be interpreted as a load for each restraint considered in isolation and ignoring the global effects(i.e ignoring the truss effect)

Cheers

Thanks

 

[OzEng80]

Yeh the code provides 'no guidance' in a lot of areas... This has a come up a couple of times recently and we have adopted the same conservative approach that you are advocating ie 2.5% to each beam with each restraint acting in the same direction. BA's Canadian reference with the alternating restraint loads is the first i've heard of the approach. Any comments on the 'sine wave' deflected shape? I'm thinking someone much wiser would be able to fill in the blanks?

 

[apsix]

hokie

I did use the term 'relatively flexible' previously.
Of course it depends on the truss geometry and span.

George's problem has 2 x 900WB's with an 18m span and horizontal truss depth (width) of 1.1m. The beams are working at close to section capacity. It is relatively flexible compared to a concrete floor diaphragm, but perhaps not compared to a roof.

I'm not saying that a truss or roof is a sub-standard bracing method but I do believe it would be folly to reduce the required lateral strength by using alternating restraint force directions. At best you could ignore every 2nd restraint force, but I wouldn't do it personally.

The initial crookedness etc and residual stresses largely influences the magnitude of the restraint force, surely it would also influence the direction.

 

[hokie66]

How deep would a solid horizontal member need to be to brace the top flange of a 900WB? What if the flange were braced with a continuous channel? What force/metre would you use in that case? I don't know, but it would be the same as for the truss. It seems to me that a 1100 deep truss is more than enough to brace a beam that is ONLY 900 deep.

 

[george69]

HI all
Not sure if this is worth anything and I have yet to read it. Unfortunately from Googlebooks so it is not complete. May trigger some thoughts but it would appear to me that the forces could act in the same direction.

Book is STeel Structures: Design and Behaviour by Salmon Johnson and Malhas

http://books.google.com.au/books?id...al bracing forces to prevent buckling&f=false

 

[asixth]

The section capacity of a 900WB175 (W36x135) is 2025kN-m (1490kip-ft) and the design actions are close to section capacity. With a moment modifier of 1.0, the unrestrained effect length cannot exceed 1,000mm (3.5') considering load height multipliers to the effective length. So at peak moment regions you need to provide restraint at 1,000mm centers which is what you are trying to achieve. To provide lateral restraint you are creating a truss out-of-plane using the flanges of the 900WB with a depth of 1,100mm (4'). So the out-of-plane truss will most likely be stiffer in the horizontal direction than what the two 900WB's are in the vertical direction. So on that basis I would say that the 1,100 deep truss will easily provide the 2.5% flange force to achieve restraint. It would be analogous to checking a W12 (310UB) bending about it's minor axis but buckling about it's major axis.

If the element providing the lateral restraint wasn't a 1,100mm deep truss than the restraint question would be a different story. If you have two beams side by side both subject to the same stress and you are relying on the two beams to restrain the other from buckling about their minor axis. What is to say that the two beams don't buckle in the same direction out-of-plane?

 

[paddingtongreen]

"If the element providing the lateral restraint wasn't a 1,100mm deep truss than the restraint question would be a different story. If you have two beams side by side both subject to the same stress and you are relying on the two beams to restrain the other from buckling about their minor axis. What is to say that the two beams don't buckle in the same direction out-of-plane?"

But you only have to prevent lateral movement of the top flange, or, rotation of the beam; both are necessary for lateral torsional buckling to take place.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[TXStructural]

If you tie the flanges together, and also provide bracing from a top flange to the adjacent bottom flange (think "Z" in section), you have eliminated LTB potential. That seems to be the simplest approach.

The horizontal truss analogy also works, since it limits the lateral movement of any portion of top flange relative to the remaining flange. For the midspan, top flange to roll (individually or separately), the truss would have to fail, or the entire beam would have to roll.

 

[george69]

Giving it some thought I think I have got a way to prove or disprove if the truss works by using an analysis program.

Creating the model:

1) Create the truss(say a plan view of it).
2) Model the top flanges of the 900WB as the chords.
3) Model the web members (I have angles for bracing)
4) Ensure out-of plane buckling of the truss is is prevented. We only want to see in-plane effects!!
5) Apply chord compression loads to both top and bottom chords that match M/D along the beam.
6) Run and elastic critical buckling analysis and see if the truss "as a whole" buckles laterally, OR, if individual chord members buckle laterally.
7)If the whole truss buckles laterally, clearly the bracing is ineffective.
8) However, if the individual chord members buckle laterally (that is sinusoidally between bracing points), then we prove that the truss IS effective in preventing lateral buckling of the entire top flange.


Clearly, the method above does not make use of lateral forces whether in the same direction or opposing directions. Would like to hear some views on the proposal.

 

[BAretired]

asixth,

What does the 175 in 900WB175 mean? If it is the weight in kg/m then it isn't the same as W36x135.

But assuming it is a W36x135, I find Mr = 2600 kn-m when its length is less than Lu (3720 mm). It has Mr of 2540 kn-m at a spacing of 4m, so on what basis do you say:

With a moment modifier of 1.0, the unrestrained effect length cannot exceed 1,000mm (3.5') considering load height multipliers to the effective length.

george69,

I believe your model will work fine to prove the adequacy of the truss, but I'm not sure that the wavelength of the beam buckling will be one meter. Unless I looked at the wrong WF section, you do not require lateral bracing as close as one meter. You could go as far as 3.7m between braces (let's say 3m). If the truss and each beam flange have identical slenderness ratios local buckling will occur more or less simultaneously with overall buckling. For that condition with a 3m spacing of braces, the truss needs to have a moment of inertia 6 times Iy/2 of each beam. I am quite sure without doing any calculation that you exceed that by far.

I believe that, in addition to the truss between top flanges, you should provide diaphragms between the two beams at each end and at the third points. Keeps everything square.

BA

 

[asixth]

BA

I was doing the calculations based on standard Australian sections and the Australian Steel Code AS4100. So all the metric calculations would have been based on the 900WB175. I suggested the W36x135 because it had similar properties as a 900WB175 just slightly larger. By doing this I was trying to open the post up to the US engineers in anticipation for a greater response. The fact that your calculations are yielding slightly higher capacities and longer unrestrained lengths is justified by the fact that a W36x135 has greater Ixx and Iyy values.

With my quote regarding moment modifiers and load height factors. A moment modifier of 1.0 would mean using a Cb value 1.0 as quoted in AISC. This is referenced in AS4100 as an ?m value.

The load height factor is a multiple added to the effective length calculations to account for the position of load relative to the shear center of the section. It is referenced in AS4100 but I am unsure whether it is used in any other steel code. For gravity loads which are applied to the top flange (above the shear center), the effective length is increased by a factor of 1.4 to account for the destabilizing effects. Not sure what research this provision from the Australian codes is based on.

Just a side note, how do you get the quote box to show?

 

[BAretired]

asixth,

Thanks. I agree that the position of the load above or below the shear center is an important variable. The NBC does not take it into account. I'm not sure about AISC.

To get this:

The quote box

You type:
[quote)The quote box[/quote]
except that you replace the curved bracket with a square one.

You can get all kinds of other interesting things by clicking on "Process TGML"


BA

 

[BAretired]

george69,

I incorrectly mentioned the ratio of moments of inertia. I should have said radius of gyration. With bracing spaced at one meter, the truss would need a radius of gyration 18 times the r[sub]y[/sub] of the beam in order to have simultaneous local and global buckling.

The radius of gyration of your truss is approximately 778mm and r[sub]y[/sub] for a W36x135 is 60.7mm (it may be different for your beam). The ratio is 12.8, less than 18, so I'm guessing the wavelength of the buckling curve would be 2m.

BA

 

[apsix]

george69
Does the critical buckling analysis allow for initial imperfections? Attached is discussion from Trahair et al regarding real vs. ideal beams.

 

[BAretired]

The radius of gyration for the horizontal truss is 1100/2 = 550mm (not 778 as stated earlier). L/r is 32.7. The Euler buckling stress for the truss with load P[sub]cr[/sub] at each end is [σ][sub]cr[/sub] = [π][sup]2[/sup]/(L/r)[sup]2[/sup] or 1843MPa.

This is beyond the elastic range which means the Euler formula is not applicable. For L/r = 33, CISC indicates a unit factored compressive resistance of 291MPa for Fy = 350MPa or 83% of yield stress.

If the compression follows the compression in the flange, a parabolic variation, the maximum value at midspan will be higher, perhaps in the order of 90% of yield stress, corresponding to kL/r of 1 to 10.

The truss is not going to buckle globally. The top flanges will buckle locally and the brace forces will alternate in direction.

BA

 

[OzEng80]

BA can you explain why the brace forces alternate direction?

Thanks

 

[BAretired]

Yes, OzEng80, I can. It is a matter of statics. Fig 2-7 attached indicates a member with braces at each end and at the third points. If the buckled configuration is as shown in the diagram, the brace forces must alternate in direction in order to satisfy statics.

BA

 

[OzEng80]

Thanks BA. That's the reference I was looking for.

 

[george69]

Hi apsix

Actually I I have startted the model in Strand 7 and will be adding in a imperfection. I'm going to have a re-read of Trahairs book in the morning.

CHeers