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Effective Length

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ToadJones

Structural
Jan 14, 2010
2,299
sketch?
 
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Am I missing something? This design 101. 1.5 m.
 
in plane, unb length = 1500
out of plane, 9000
 
Toad: He mentioned that it is braced at 1.5m in both directions.
 
I wasn't following him totally.
Yes, I'd say 1500 for all.

 
Agreed - I would normally treat it as 1.5m also. However, my vertical column / truss is actually a scaffold subject to the scaffold code BS EN 12811 and guidance document from that TG20:08. In these they state if the structure is 6m high with 2m lifts, restrained top and bottom as my exmaple, then the effective length is 4m???

Doesn't make sense to me but makes a huge difference in leg capacity.
 
I disagree with 1500, with the all of the diagonals except one in the same direction, any vertical load will compress the diagonals and will force a bowing of the column.

The 4.0M may refer to the overall column. A latticed column is designed as a unit and as individual ties and struts.

Michael.
Timing has a lot to do with the outcome of a rain dance.
 
Not sure how to put this, but if I am designing a lattice column, I use FEM with the individual members with K=1.0 and unbraced length = spacing between braces so long as the braces are stiff enough to provide lateral restraint.
The FEA will take into account the column shortening and relative stiffness of the members and how much load will go from the vertical columns into the diagonals.
If you have a massive W-shape column with light angle lattice work, not as much vertical load goes into the bracing as opposed to if the lattice bracing was a very heavy section.
For the most part, the heavy lattice columns I work with are latticed for lateral loading.
 
For the chords, Lbxx = Lbyy = 1.5m
For the lattice beam as a whole, Lbxx = Lbyy = 9m

FEA will deal with this as a 2nd order, P-Delta effect.
 
@Toad, I was assuming scaffolding in this case, and thinking that the braces might be the same or similar weight as the columns.

It seems to me that the overall column could be less than full height, the overall base is, to all intents and purposes, fixed. The column bases are free to move up but they don't want to, they cannot move sideways or down, so I think moment and gravity loads for the overall column are resisted.

Michael.
Timing has a lot to do with the outcome of a rain dance.
 
In a truss with members (pin connected) the effective length for each element is that element length. However, I don't know if there is specific provision for scaffolding.
 
How is it anything other than 1500?

paddington - I recognize your point about the brace compressing and dragging the column in a little bit, but I don't believe that effects the buckling length of the column. The brace is still there and would prevent the buckling mode of anything greater than 1500. The buckling mode of the 1500 segment may be very slightly out-of-plumb as a result of the diagonal compression that you mention, but I don't believe it changes the buckling length.
 
I probably wasn't terribly clear, but when I said 1500, it was for the individual vertical compression members only, not for the entire latticed column.
 
Are we saying that the scaffold can never ever buckle as a system?

However, I agree that for member design the unbraced length will be 1.5m with a k=1.0. But if we provide each member to 3 times its required strength and avoid member buckling as a failure mode, can't the system buckle globally? Just a thought....

 
slick-

No, I agree the whole system can buckle. That's how you would check a latticed column. My comment was for the the individual vertical members.

I just have a really hard time believing that the global buckling would control the design for all but very tall pieces. I haven't put pencil to paper, but it makes intuitive sense to me. I, and, consequently, r increase as exponential functions of the distance between the individual vertical pieces. I could be way off base, but that's what makes qualitative sense to me.
 
Agree with others that individual brace length is 1.5m however for overall I would use AISC E6 (built up members) to determine the (KL/r)m which is a modified (KL/r) for use in equations in E3 and/or E4.

EIT
 
Gents - thanks for your comments - I think I get the jist of it. I've done some homework and the answer is specific to scaffold as it is not a tower but actually a facade (a long tower). What we do is tie every other standard back to the building and then the ledgers (horizontals) stabilise the in between members. This effective length quoted is from analysis and represents a worst global case - I.E. a tied standard might be 1.5m but the adjacent might be 4m as it is not tied and only stabilised by the adjacent ledgers - hence they quote the 4m for structural use.

This makes sense to me now - but I thought I was going mad!
 
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