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Twin beam lateral restraints at top flange

Twin beam lateral restraints at top flange

Twin beam lateral restraints at top flange

(OP)
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

RE: Twin beam lateral restraints at top flange

The bracing acts as a horizontal truss, apply your stability, and any other horizontal loads to it. Make sure that the bracing is up close enough to the top flange to be effective.

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

RE: Twin beam lateral restraints at top flange

Your code is slightly more conservative on that point than the NBC which requires 2% of flange compression in each brace unless a more detailed analysis is performed.  When a detailed analysis is performed, the bracing requirement is, in my experience, always less than 2%.

Bracing forces are acting alternately in opposite directions, i.e. if Braces 1, 3, 5 and 7 act east, then Braces 2, 4 and 6 act west.  Thus you do not include these forces in your truss design.

  

BA

RE: Twin beam lateral restraints at top flange

I have designed similar members, which were closely spaced wide flange beams, and using Vierendeel configuration cross connections.  Model the top flange as a truss for out of plane deflection resistance (stiffness of the truss is primary), using your code-required bracing forces.

Also, look at whether the loads will be eccentric, causing warping of the top flanges, and brace in cross section if needed to prevent rotation.

RE: Twin beam lateral restraints at top flange

The top flange of each beam must be connected to the truss for a force 2.5% of the compression in it, but the forces are not cumulative as you show on your sketch.  That would make no sense at all.  If you had elected to space panel points at 0.5m instead of 1m, by that reasoning, you would need to design the truss for twice as many forces of the same magnitude.  Or, if you had opted to space them at 2m, you would need to provide only half as much lateral load.

The required force, Pb in a brace can be determined in another way.  This is the expression in the NBC:

Pb =β(Δo + Δb)Cf

Pb = the force used to design the bracing system.  When two or more points are braced, the forces Pb alternate in direction.

I believe it would be conservative to design your truss for one midspan force F = 2.5% of the combined maximum compression in the two top chords plus wind pressure if applicable.

 

BA

RE: Twin beam lateral restraints at top flange

The 2.5% restraint force is required at spacings equal to the effective length which results in the member capacity you require.
If the restraints (panel points) are at closer centres the restraint force can be averaged out to achieve the same overall force. (That's how I interpret AS4100 Cl.5.3.4.1)

 

RE: Twin beam lateral restraints at top flange

Can you post a copy of the clause in question?

BA

RE: Twin beam lateral restraints at top flange

George, I misunderstood the question. That is not "K" bracing!

When you said K bracing, I thought it would be short span with a single K bracing with a strut in the middle, local and global would have been the same.

I defer to BA who didn't misunderstand.

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

RE: Twin beam lateral restraints at top flange

BA, as requested. Enjoy.

5.4.3.1 Restraint against lateral deflection The lateral restraint at any cross-section considered to be fully, partially or laterally restrained in terms of Clause 5.4.2 shall be designed to transfer a transverse force acting at the critical flange (see Clause 5.5) equal to 0.025 times the maximum force in the critical flanges of the adjacent segments or sub-segments, except where the restraints are more closely spaced than is required to ensure that M* equals Mb.
When the restraints are more closely spaced, then a lesser force may be designed for. The actual arrangement of restraints shall be assumed to be equivalent to a set of restraints which will ensure that M* equals Mb. Each equivalent restraint shall correspond to an appropriate group of the actual restraints. This group shall then be designed as a whole to transfer the transverse force of 0.025 times the maximum force in the critical flanges of the equivalent adjacent segments or sub-segments.

RE: Twin beam lateral restraints at top flange

(OP)
apsix

Thanks for posting the AS4100 reference. And you are correct, if the restraints are closer, then you don't need to design them for the full load - obviously a pro-rata and easily done.

The question, which I believe BA correctly answers via Canadian approach, is that the "restraining" forces are in opposing directions and NOT all in the one direction - which I thought. Clearly the opposing directions would eventually cancel out. Refer to the attached. I suppose, if the number of braced sections is even, there could be a nett effect in one direction.

Paddingtongreen - sorry to confuse you but I do have K bracing. The sketch I attached was to just show an example of "bracing" (i.e any bracing configuration), to the top flange.  


cheers

RE: Twin beam lateral restraints at top flange

george69,

I believe your sketch with opposing forces is more in keeping with the NBC approach for bracing members generally.

When the bracing forces are being provided by a truss in the horizontal plane and when there are other lateral forces on the truss such as wind, the lateral deflection of the truss plays a role in stability.  If the truss deflects Δ under wind pressure, for example, the top flanges become more critical in lateral buckling because of the eccentricity.  

I don't believe the clause cited by apsix contemplated that problem and I don't think NBC considered it either.  If the lateral deflection of the truss is significant, it will assuredly affect the buckling capacity of the top flanges.

Need more time to think about it.

BA

RE: Twin beam lateral restraints at top flange

(OP)
BA retired

OK - FYI In this instance there is no lateral wind. Just pure vertical point load at midspan.

As you've seen from the AS4100, it specifically refers to "THE lateral restraint" which I take to mean individual elements need to restrain 2.5% of flange force taken laterally. But it does not specify alternate directions like the canadian approach.

This is ultimately my dilema. Reading the AS4100 blindly - one could be orientating all the lateral loads in one direction and have serious lateral load at the supports to deal with. Whereas, using the canadian approach, the loads are more or less cancelled out along the span.

I'm not familiar with the canadian code, but I gather from your doubt, that clause 9.2.6 of the canadian code is for compression members that have lateral restraints? If that is the case then I believe there is no difference as the top flange, is in essence, a compression member.


I look forward to hearing your reply.


 

RE: Twin beam lateral restraints at top flange

I'm not sure that what I have said is the singular "Canadian approach".  While I do not have AISC standards at my disposal, I believe from what I have gleaned from previous posts on this forum, that AISC has a similar approach to ours.  

The requirement for lateral bracing of a compression flange comes from theoretical considerations based on an assumed initial misalignment of the member due to accidental eccentricities during the fabrication or delivery process.  These limits are specified in our code as L/1000 or L/500 or other.  Based on this rather debatable starting position, the requirement for lateral bracing may be determined using somewhat simplified assumptions first proposed by George Winter.

The value of 2% in my code or 2.5% in your code is conservative in the extreme.  It refers to the capacity of an individual connection which can act in either direction, but to add these together in a cumulative lateral load would be totally wrong.

BA

RE: Twin beam lateral restraints at top flange

George - I imagine that the alternate bracing directions is due to the sinusoidal deflected shape of the compression member (picture a buckled member over multiple supports). It would be nice if the codes were explicit enough to cover all scenarios but there should be enough here to piece together a valid design approach.

RE: Twin beam lateral restraints at top flange

The 2-2.5% value may be conservative but the idea that you can alternate the direction as a rule seems questionable to me.

If the member has an intial imperfection curvature in the horizontal plane all the restraint forces will tend to be in the one direction, especially for relatively flexible braces such as the truss.

Of course, if designing to AS4100 all the forces must be taken to be cumulative.

RE: Twin beam lateral restraints at top flange

apsix - AS4100 requires the design of a cumulated bracing force for parallel restrained members (ie a series of beams) - it doesn't give any guidance to alternating restraints 'along the beam'. It dictates that the two beams can buckle in the same direction at a restraint but nothing beyond that. I believe that as a compression member, restraints would be required on alternating sides due to the sine wave deflected shape. This seems to be supported by the Canadian code. There is probably some Euler derived formula that describes this?

RE: Twin beam lateral restraints at top flange

OzEng80
The fact that there is no guidance isn't a reason to take an overly optimistic view.

If the restraints are provided by a roof or floor diaphragm it doesn't make any real difference.

If you are relying on a flexible restraint, such as the truss discussed here, it would be brave to assume alternating directions.

RE: Twin beam lateral restraints at top flange

apsix,
Why would you consider a horizontal truss to be a flexible restraint?  I can't see the alternating direction approach, but the truss would provide bracing approaching that of a diaphragm.  Think of how it would compare with typical roof bracing for the compression flange of rafters.

RE: Twin beam lateral restraints at top flange

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?

RE: Twin beam lateral restraints at top flange

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.

RE: Twin beam lateral restraints at top flange

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.

RE: Twin beam lateral restraints at top flange

(OP)
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=Jnhsmzm_EJoC&;pg=PA460&lpg=PA460&dq=lateral+bracing+forces+to+prevent+buckling&source=bl&ots=KGCTP4wTLY&;sig=DYa6M1YL-xIH00JL6t_xV0fRvbo&hl=en&ei=ZK2QS7GkLNGGkAXiorCADQ&sa=X&oi=book_result&ct=result&resnum=2&ved=0CAkQ6AEwATgK#v=onepage&q=lateral%20bracing%20forces%20to%20prevent%20buckling&f=false

RE: Twin beam lateral restraints at top flange

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?

RE: Twin beam lateral restraints at top flange

"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.

RE: Twin beam lateral restraints at top flange

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.

RE: Twin beam lateral restraints at top flange

(OP)
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.
  

RE: Twin beam lateral restraints at top flange

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:

Quote:

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

RE: Twin beam lateral restraints at top flange

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 αm 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?

RE: Twin beam lateral restraints at top flange

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:

Quote:

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

RE: Twin beam lateral restraints at top flange

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 ry of the beam in order to have simultaneous local and global buckling.

The radius of gyration of your truss is approximately 778mm and ry 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

RE: Twin beam lateral restraints at top flange

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 Pcr at each end is σcr = π2/(L/r)2 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

RE: Twin beam lateral restraints at top flange

BA can you explain why the brace forces alternate direction?

Thanks

RE: Twin beam lateral restraints at top flange

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

RE: Twin beam lateral restraints at top flange

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

RE: Twin beam lateral restraints at top flange

(OP)
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  

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