Bracing of Beams against lateral buckling
Bracing of Beams against lateral buckling
(OP)
My question relates to the CSA S16.1-94 steel design code, Clause 20.2 (I know that there is a later Standard but I do not have it here at home).
The formula kb = (beta Cf / L) x (1 + do / db),
where L is the length between brace points, and beta is a factor that increases with the number of equally spaced braces. It is 2, 3, 3.41 and 3.63 for 1, 2, 3 or 4 equally spaced braces, respectively.
The force in the brace equals kb db. Sorry I don't have the alt codes handy for the Greek letters.
This seems to mean that the closer the braces are spaced, the smaller L becomes, the larger kb becomes and the larger the force in the brace becomes (do and db are based on the length of the member that is being braced). How can this be? I would have thought that the closer the braces are spaced, the smaller the force in each brace.
The formula kb = (beta Cf / L) x (1 + do / db),
where L is the length between brace points, and beta is a factor that increases with the number of equally spaced braces. It is 2, 3, 3.41 and 3.63 for 1, 2, 3 or 4 equally spaced braces, respectively.
The force in the brace equals kb db. Sorry I don't have the alt codes handy for the Greek letters.
This seems to mean that the closer the braces are spaced, the smaller L becomes, the larger kb becomes and the larger the force in the brace becomes (do and db are based on the length of the member that is being braced). How can this be? I would have thought that the closer the braces are spaced, the smaller the force in each brace.






RE: Bracing of Beams against lateral buckling
Pb = β[Δo + Δb]Cf/L
Since Δo is a function of L (Δo = 0.002L), the required brace force is not getting larger with smaller spacing beyond four equally spaced braces.
You can find Greek letters and other symbols if you click on the Ω icon in the line above"Submit Post".
BA
RE: Bracing of Beams against lateral buckling
So following this example, for braces at third points of a member there would seem to be a larger force per brace than a single brace at mid-length? What am I doing wrong?
Thank you for the tip on how to find the symbols on here. That makes things much easier!
RE: Bracing of Beams against lateral buckling
My question has been asked before, and the formula does give a greater force per brace when 2 braces are used than when 1 brace is used. That is correct, because with a single brace you are trying to force the member into 2 half sign waves, but with 2 braces you are trying to force the member into 3 half sign waves and that requires more force per brace than to force it into 2 half sign waves.
RE: Bracing of Beams against lateral buckling
BA
RE: Bracing of Beams against lateral buckling
RE: Bracing of Beams against lateral buckling
In the case of only one brace at L = 2500, Δo is of course 5mm because 2L = span.
BA
RE: Bracing of Beams against lateral buckling
RE: Bracing of Beams against lateral buckling