ajk1
Structural
- Apr 22, 2011
- 1,791
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.